Download Problem 1-2 - IPN-Kiel

47. International
Chemistry Olympiad
Azerbaijan 2015
National German
Competition
Volume 21
Chemistry Olympiad 2015
National German Competition 2015, Volume 21
Translated and published by Wolfgang Hampe
Contact addresses:
IPN University of Kiel, z.H. PD Dr. Sabine Nick
Olshausenstraße 62
24098 Kiel
tel: +49-431-880-3116
fax: +49-431-880-5468
email: [email protected]
IPN University of Kiel, z.H. Monika Barfknecht
Olshausenstraße 62
24098 Kiel
tel: +49-431-880-3168
fax: +49-431-880-5468
email: [email protected]
Wolfgang Hampe
Habichtweg 11
24222 Schwentinental
tel:
+49-431-79433
email:
[email protected]
Association to promote the IChO
(Association of former participants and friends of the IChO)
Internet address:
www.fcho.de
This booklet including the problems of the 47th IchO and the latest statistics is available as
of September 2015 from http://www.icho.de ("Aufgaben")
4
Chemistry Olympiad 2015
Contents
Part 1: The problems of the four rounds
First round
Second round
Third round, test 1
Third round, test 2
Fourth round, theoretical test
(problems solved at home) ....................................
(problems solved at home) ....................................
(time 5 hours)..........................................................
(time 5 hours)..........................................................
(time 5 hours)..........................................................
6
10
18
24
31
Fourth round, practical test
(time 5 hours) .........................................................
42
Part 2: The solutions to the problems of the four rounds
First round
Second round
Third round, test 1
Third round, test 2
Fourth round, theoretical test
.................................................................................
.................................................................................
.................................................................................
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.................................................................................
47
53
60
66
72
Part 3: The problems of the IChO
Theoretical problems
Practical problems
Solutions
.................................................................................
.................................................................................
.................................................................................
81
95
110
Part 4: Appendix
Tables on the history of the IChO
............................................................................
120
3
Chemistry Olympiad 2015
4
Problems
Part 1
The problem set of the four rounds
5
Problems Round 1
First Round (homework)
Problem 1- 1
Quite Old!
Natural carbon consists of three isotopes.
a)
Which are these isotopes? How do they differ in their atomic composition?
One of these isotopes is radioactive with a half-life of t½ = 5730 years.
b)
How is this isotope generated in nature? Give a formation equation!
The radioactive carbon isotope is an  emitter.
c)
Write down the decomposition equation.
d)
What is the law for the radioactive decay?
e)
What is the meaning of "half-life" of an isotope? Derive a general formula for the half-life, starting with the law for the radioactive decay.
f)
After which time does radioactive material stop to disintegrate?
g)
What is the influence of temperature, pressure and similar conditions on the rate of radioactive
decay?
Radiocarbon dating is an important method to determine the age of carbon containing material.
h)
Explain this method! Give the reason why examined materials have to be former living objects or
their derivatives.
The paper of a treasury map found in 2013 has a decay rate of 14.48 decays per g of carbon per minute. In natural carbon this decay rate is 15 decays per g of carbon per minute.
i)
Calculate the age of the map? Should the finder go on a treasury hunt?
j)
Account for the fact that the age of bones of dinosaurs cannot be determined by the method of
carbon dating.
Problem 1-2
Fats and Oils
Fats and oils are esters of carboxylic acids with glycerol. They can be solid, semisolid and liquid. Fats
which are liquid at room temperature are called oils.
Depending on their origin fats have a characteristic composition of fatty acids.
a)
Explain what "characteristic composition of fatty acids" means and show such a characteristic
for an example chosen by you.
There is a simple code to characterize fatty acids by giving the proportion of the total number of
carbon atoms (m) and the number of double bonds (n) (as it is done in sport results). The code for
oleic acid is 18:1.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
O
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Fig. 1: Structure of oleic acid
6
OH
Problems Round 1
b)
Write down the empirical formulae and plot the structures of the fatty acids below. Give their
code.
i) Capric acid ii) Linoleic acid
iii) Linolenic acid iv) Erucic acid
Natural fats which contain (poly) unsaturated fatty acids are especially valuable for dietary intake.
c)
Explain by considering intermolecular interaction why saturated fats are solid while unsaturated
fats are rather liquid.
More than 90 % of fats produced worldwide serve as foodstuff for men or animal feed, a small part
is raw material for producing other chemicals. There are special regions in a fat molecule where a
chemical attack is possible.
OH
O
OH
O
OH
O
O CH2
O CH
O CH2
Fig. 2: Structural formula of ricinus oil
d)
Mark the regions where a chemical attack is possible and give a name to it.
A possible reaction of a fat is the saponification with a solution of sodium hydroxide to give glycerol
and soap.
e)
Give the reaction equation of the saponification of a fat (using structural formulae). Show the
detailed mechanism of the saponification of one of the three fatty acid residues. Explain why this
equilibrium reaction drifts to the products.
The saponification of copra oil needs several hours while the saponification of 3-nitrobenzoic-acid
methyl ester is finished after only ten minutes.
f)
Account for this fact.
Soaps belong to the tensides.
g)
What characterizes a tenside? Explain the cleaning effect of tensides.
h)
To which kind of tensides do soaps belong? Which other kinds of tensides do exist? Give an example to each kind and draw the structural formula.
Problem 1-3
Oxygen ...
Oxygen is the most frequent element on earth having a mass ratio of 49.4 %.
a)
Give five natural deposits of oxygen.
The production of oxygen from air is technically performed by a special procedure in Germany
known as Linde operation.
b)
How can oxygen be produced in a laboratory? Write down the names of three methods and the
equations of the reactions.
7
Problems Round 1
Besides oxygen nitrogen (consisting of N2 molecules) is the major component of air
c)
Are there differences between O2 und N2 molecules concerning their magnetic properties? Account for your answer using MO diagrams.
Oxygen molecules (O2) are denoted as triplet oxygen. Moreover there is a more reactive species, the
so called singlet oxygen. These names arise from the total spin multiplicity M which can be calculated with the formula M = 2·S + 1 (S = total spin).
d)
Account for the names of these two "kinds" of oxygen using this formula!
In text books you often find the Lewis structure for dioxygen (O2)
shown on the right hand side.
e)
O O
Lewis structure of O2
Does this Lewis structure show the correct distribution of electrons in the O2 molecule? Account
for your answer.
Problem 1-4
... and Oxides
Oxygen forms binary compounds with nearly all other elements.
a)
From which elements no isolatable oxygen compound is known until now?
b)
Write down the equations of the reactions of oxygen with
i) White phosphor
ii) Yellow sulfur
iii) Lithium
iv) Calcium
Which acid/base properties do the aqueous solutions of the products show?
In its covalent compounds oxygen has in most cases the coordination number 1, 2 or 3.
c)
Give one example for each of these three coordination numbers plotting its Lewis structure.
Besides some exceptions the oxidation number –II is assigned to oxygen.
d)
Give four compounds in which oxygen has another oxidation number than –II. Do not name
more than one compound for each different oxidation number.
In quantitative analysis oxides can be used to determine the amount of other elements for example
iron, cobalt, nickel, tin and aluminum.
To determine the iron(III) content in a solution it is precipitated with ammonia, filtered through ashfree filters, washed with water and at the end with ammonium nitrate solution.
The filter with the precipitate is given into a porcelain crucible
and heated with a Bunsen burner, at first slowly and then up to
maximal 700 °C. You have to pay attention that there are no
reduction processes and that besides Fe2O3 no Fe3O4 is formed
which would falsify the result of the analysis.
A sample of iron(III) chloride is weighed into a 250.0 mL measuring flask which then is filled with
water up to the calibration mark. Three samples of 50.0 mL each are taken and treated as described
above.
Weight of the ash: 0.2483 g / 0.2493 g / 0.2488 g.
e)
8
i)
Write down the equation for the formation of iron(III) oxide from iron(III) hydroxide.
Problems Round 1
ii)
What is the reason for washing with ammonium nitrate? Why should ammonium chloride
not be used in the last step? Account for your answer.
iii) Why is ash-free filter paper used?
iv) With which simple physical measurement could the unwanted formation of Fe3O4 be detected?
v) Does the formation of Fe3O4 lead to a higher or to a lower calculated content of iron? Account for your answer.
vi) Calculate the mass of the iron(III) chloride sample which was given into the measuring flask.
Problem 1-5
... another Oxide - An Elaborate Determination
290.0 mg of a metal oxide (MeO2), concentrated hydrochloric acid as well as a solution of potassium
iodide are placed in an apparatus as shown below.
Fig. 3: Apparatus for a quantitative determination of a metal oxide
The hydrochloric acid in the dropping funnel is pressed into the reaction vessel A using carbon dioxide. This mixture is heated to maintain light boiling for 30 minutes.
During this time the apparatus is continuously purged with small amounts of carbon dioxide to remove the volatile content of vessel A.
When the reaction in vessel A has finished the content of the washing flask C is quantitatively transferred into the Erlenmeyer flask B.
The content of the Erlenmeyer flask B is titrated with a standard solution of sodium thiosulfate (c =
0.1 mol /L) until the mixture is nearly discoloured. Then some drops of solution of starch are added
and the titration is continued until the solution is totally discoloured.
24.25 mL of the thiosulfate solution are consumed for the titration.
a)
Write down reaction equations for all reactions in the apparatus as well as for the reactions
during the following titration.
b)
Why are drops of starch solution added?
c) Account for the reason of such an elaborate procedure. What is the influence of air and water
(moisture) on the accuracy of the determination?
d)
Determine the metal in the oxide.
9
Problems Round 2
Second Round (homework)
Problem 2-1
Afraid of Water!
The nitrate of M, dissolved in a small amount of water at 50 °C, is added to a 7.5-fold excess of a
warm (also at 50 °C) aqueous solution of X. A solid Y forms which converts to compound Z · n H2O
when cooled down. The mass ratio of the metal M in Z amounts to 33.04 %.
The following information is given:
 The metal M dissolves is nitric acid of the concentration of c = 8 mol/L but not in sulfuric acid
or hydrochloric acid of the same concentration.
 The metal M dissolves in a hot solution of sodium hydroxide.
 Finely dispersed M ignites in air.
 If a solution of sodium hydroxide is added to an aqueous solution of M a precipitate forms
which dissolves in an excess of sodium hydroxide solution.
 M in an aqueous solution does not form a complex compound with ammonia.
 When dissolving X in water the solution cools down.
 Y dissolves slightly in water.
 As a solid Y has a layer structure.
 The conductivity of solid Y rises when heated.
 Y shows thermochromism.
 Z · n H2O can be dehydrated in a drying oven at 150 °C.
 Z and Z · n H2O decompose in water.
 When Z · n H2O is heated a gaseous compound G is formed besides water vapor and a metallic mirror of M is formed
 X and G react in an aqueous solution with each other to yield compound H.
a)
Determine M, X, Y and Z. Account for your decision.
b)
Write down the reaction equation for the two step formation of Z using the compounds mentioned above.
c)
Determine H. Write down the equation of its formation from X and G! Draw the Lewis structure
of H. Which structure do you expect according to the VSEPR model. Account for your answer.
Mass percentage
The water containing compound Z · n H2O can be dehydrated in a drying oven. The image shows the
thermography (TG) plot.
Temperature/°C
10
Problems Round 2
d)
Determine the number of water molecules n (n integer) in one formula unit of Z · n H2O using
the observed decrease of mass.
Beside Z another compound was found, which consists of the same elements with the same oxidation numbers as in Z. The mass ratio of the metal M, however, has only the value of 26.13 %.
e)
Give the empirical formula of this compound.
f)
Give at least three compounds which lead to a decrease of temperature when dissolved in water. Give an explanation of this phenomenon!
g)
Explain the formation of the metallic mirror when Z · n H2O is heated. Write down the reaction
equation and apply oxidation numbers.
h)
Give the equation for the reaction of an aqueous solution of the nitrate of M with an excess of
an ammonia solution.
Problem 2-2
Organic Chemistry – Short and Crisp
5-Methylfuran-3-on (2) is needed as a reagent for the synthesis of a natural compound. It can be
produced in four steps starting with acetylacetone (1):
O
O
1
Hexan,
EtEt33,N,
hexane,
TMSCl, RT,
48 h
A
NBS,
i) LDA, THF,
NBS, DMC
DCM,
reflux
Rückfluss,
-78 °C,
B
2h
30 min
C11H24O2Si2
ii) TMSCl,
0 °C, 1 h
C
Et2O,
K2CO3,
RT, 12 h
O
O
2
Hints:
- In the reaction of 1 to A two isomers form.
- In the 1H-NMR spectrum (A in CDCl3 at 298 K) of the mixture of isomers of A six singlets are
found. The intensities are written below the signals, the chemical shiftabove the signals.
11
Problems Round 2
-
a)
b)
c)
d)
e)
f)
g)
In the mass spectrum of C the following values for the molecular peak m/c are found (intensities
in brackets):
177.96 (100.0%), 179.96 (97.3%), 178.97 (5.6%), 180.96 (5.3%)
Draw the structural formulae of A, B and C and show how you derived the structures from the
hints and the reaction equation.
Which kind of isomers are formed in the reaction 1  A? Draw the structural formulae!
Analyze the 1H-NMR spectrum of A: Assign all hydrogen atoms of b) to the corresponding signals. Determine the ratio of the two isomers of A. Calculate the intensities of the different signals using the ratio of isomers.
Why do you find two peaks with approximately equal intensities at 177.96 and 179.96 in the
mass spectrum of C? Explain!
In the first step of the synthesis of 1  A triethylamine, Et3N, is used. What is it needed for?
Explain!
In the second step of the synthesis of A  B lithium diisopropylamide, LDA, is used. Is it possible
to use Et3N here analogue to the reaction of 1  A? Account for your answer.
Would a reaction of acetylacetone with N-bromosuccinimide, NBS, in the presence of a base lead
to C, too? Why is the way via A and B used in the synthesis? Give an explanation!
In an aldol reaction followed by an elimination of H2O product 2 is linked to another structural element. The unit 3 is formed which you find in the natural compound, too.
In the natural compound solved in a deuterized solvent an H/D exchange is observed which can be
explained by keto-enol tautomerism.
h)
Show a mechanism of the H/D exchange at the methyl group of the structural unit 3. How many
of H/D exchange reactions are possible?
Problem 2-3
Hückel Theory
The Hückel theory can be used to describe π electron systems qualitatively. The p orbitals which are
involved in the π bonds (per definition pz orbitals) have to be taken into consideration. The π molecular orbitals (MO) are linear combinations of them. By doing so bonding and antibonding MOs are
formed. If they are accommodated similarly a non-bonding interaction occurs. A simple example is
ethene (Fig. 2).
The energy levels of ethene result from bonding and antibonding interaction and contain the empirical parameters  (energy of the electron in the isolated atom) and β (coupling of the atom orbitals in
the molecule, β > 0).
The total π energy of ethene results from the occupation of the energy levels with two electrons as
follows
E = Σi ni εi = 2 (α + β) + 0 (α – β) = 2 α + 2 β,
with ni = number of electrons in MO i and εi = associated energy of the MO.
12
Problems Round 2
antibonding
bonding
Fig 2: π Molecular orbitals of ethene following the Hückel theory
A Hückel MO scheme can also be established for cyclic and planar π systems. There is a simple
“trick” for working out the orbital energies: Frost-Musulin diagrams. Starting from the energy level
of the atom orbitals () a circle with the radius 2β is drawn. The molecular framework of an n-cyclic
system is then drawn as a polygon into the circle with one atom put at the bottom. The atomic positions then map on to the energy level diagram of the π system. Fig. 3 shows the Frost-Musulin diagram of benzene.
Fig. 3: Frost-Musulin diagram of benzene
If you compare the π bonding energy of benzene with that of hexatriene you find stabilization by
delocalization of 1.0β. This is often interpreted as the so called aromatic stabilization energy: Benzene gains additional bond energy and is called an "aromatic" compound. Anitaromatic compounds
do not show such an effect, they are unstable.
a)
Design the Frost-Musulin diagram of a planar 4- and 7-cyclus. Give the charge of the molecule
for different electron configurations (4-membered ring: 2, 4 and 6 π electrons; 7-membered ring:
6 and 8 π electrons) and calculate the stabilization β compared to the open chain versions of the
molecules (ions). Are these compounds (ions) aromatic?
(The π energy levels of the aliphatic versions are:
Butadiene:  ± 1,618 · β;  ± 0,618 · β;
Heptatrienyl cation: ;  ± 1,848 · β;  ± 1,414 · β;  ± 0,765 · β).
b)
Which are the conditions a compound has to fulfill in order to be called aromatic?
13
Problems Round 2
c)
Indicate which of the following molecules and ions are aromatic and which are not. Account for
your decision by using the conditions of b)i) Pyrrole ii) Allyl anion iii) Azulene iv) 1H-Pyrrolium cation v) Pyridinium cation vi) Caffeine.
The shape of the wave function of a special energy level can be found by linear combination of the pz
orbitals. The result for the cyclopentadienyl anion (Cp–) is shown in fig. 4:
Fig. 4: Frost-Musulin diagram of cyclopentadienyl anion with the MOs
These MO diagrams are of great importance in coordination chemistry. The coordination mainly
takes place by HOMO (highest occupied molecular orbital) - LUMO (lowest unoccupied molecular
orbital) interaction.
d)
Write down the HOMO and LUMO of the cyclopentadienyl anion!
An iron(III) cation is coordinated by a cyclopentadienyl anion (Fig. 5).
Fig 5: Orientation of the η5-cyclopentadienyl iron(III) cation
e)
Determine the interactions between the valence orbitals of the iron cation (3d, 4s und 4p) and
the frontier orbitals of the cyclodienyl anion. Use the orientation in the coordination system as
shown in fig. 5. Record your results in a table as shown below (x = interaction expected, – = no
interaction expected). Account for your assignment by one example for HOMO and LUMO each
with the help of orbital plots which illustrate the interaction.
Orbital Fe
3dx2-y2
3dz2
...
HOMO (Cp-)
LUMO (Cp-)
In ferrocene two Cp- ligands are coordinated to one iron(II) cation. In this complex iron has 18 valence electrons and thus a stable noble gas configuration. This "magic" number is found in lots of
coordination compounds.
14
Problems Round 2
f)
Which compound is more stable: [Rh(η5-Cp)2] or [Ru(η5-Cp)2]? Which redox property should the
less stable compound have according to the configuration of the valence electrons at the metal
center? Account for your answer!
g)
The complex [(η5-Cp)2Ru2(CO)4] may have three diasteromeric structures. Draw 3-D structures
and give the number of valence electrons at each metal center.
Problem 2-4
Disproportionation of Copper
0.168 g of copper(II) nitrate is dissolved in water to yield 100 mL solution. The pH of this solution is
4.4.
a)
Why does an aqueous solution of copper(II) nitrate react acidic? Write down the reaction equation! Calculate the pKa value of the first step of protolysis.
b)
Determine the pH from which copper hydroxide precipitates in a solution with c(Cu2+) = 1.03·10–2
mol/L. (Ksp, 25°C = 1.6 · 10–19)
There are two redox equilibria for Cu+ ions:
Cu+ + e–
Cu
Eo1 = 0.52 V
Cu2+ + e–
Cu+
Eo2 = 0.16 V
c)
Write down the reaction equation for the disproportionation of Cu+ ions and calculate the equilibrium constant for this reaction at 22 °C.
d)
Which oxidation state of copper ions should be the most stable according to its electron configuration? Which one is the most stable in an aqueous solution? Account for your answer.
e)
10 mmol of copper(I) nitrate are dissolved in 1 L of water at 22 °C. Calculate the composition of
the copper containing species in mol/L. (Use K =1.72 · 106 instead of the result in c). )
Copper(I) oxide is suspended in a copper(II) solution of c(Cu2+) = 0.01 mol/L at 22 °C.
(Ksp (CuOH) = 1.0 ·10–15)
f)
Calculate the pH from which copper(I) oxide is stable in an aqueous solution. Which influence
does the temperature have in the range from 0 °C to 100 °C? Draw a graph!
15
Round 3 Test 1
Problems Round 3
The top 60 of the participants of the 2nd round are invited to the 3rd round, a
one-week chemistry camp.
Test 1
Test 2
Göttingen 2015: Problems 3-01 to 3-09
Göttingen 2015: Problems 3-11 to 3-20
time
your name
relevant calculations
atomic masses
constants
answers
draft paper
problem booklet
5 hours.
write it on every answer sheet.
write them down into the appropriate boxes.
otherwise you will get no points
use only the periodic table given.
use only the values given in the table.
only in the appropriate boxes of the answer
sheets, nothing else will be marked.
use the back of the pages of the problem
booklet, but everything written there will not
be marked.
you may keep it.
Good Luck
16
Problems Round 3 Test 1 + 2
Useful formulas and data
G0 = H0 - T·S0
G0 = - E·z·F
G = G0 + R · T· ln Q
ln (Kp1/Kp2) =
−H0
R
G0 = - R·T·ln K
·(T1-1 - T2-1)
p·V = n·R·T
for ideal gases and osmotic pressure
Nernst equation
:
E = E0 +
R ·T
z ·F
·ln (cOx/cRed)
for metals
E = E0 +
for non-metals
E = E0 +
for hydrogen
E = E0 +
R ·T
z ·F
·ln (c(Mez+/c0)
R ·T
z ·F
·ln (c0/c(NiMez-)
R ·T
c(H+ )/c0
z ·F
(p(H2 )/p0 )1/2
·ln
with c0 = 1 mol/L, p0 = 1.000∙105 Pa
Rate laws
0. order
c
= co - k·t
1. order
c
= co· e k 1 t
2. order
c-1 = k2·t + co-1
Arrhenius equation:
k = A ∙ e-Ea/(R∙T)
A
Ea
pre-exponential factor
activation energy
Law of Lambert and Beer:
A = ·c·d
A

d
c
absorbance
molar absorption coefficient
length of the cuvette
concentration
Transmission T =
I
I0
I0
Absorbance A = lg
T = K ·
Freezing point depression
with I = intensity
I
n
m(Solvent)
n
K
Speed of light
c = 3.000∙108 ms-1
Gas constant
R = 8.314 JK-1mol-1
Faraday constant
F = 96485 Cmol-1
Avogadro constant
NA = 6.022·1023 mol-1
po = 1.000·105 Pa
1 atm
amount of particles dissolved
cryoscopic constant
= 1.013·105 Pa
1 bar
= 1·105 Pa
1 Å = 10-10 m
A periodic table was provided
17
Round 3 Test 1
Third Round Test 1
Problem 3-01
Basic Knowledge
A
a)
b)
c)
Fill in the missing numbers:
 Na2S2O3 +  I2
 NaI +  Na2S4O6
2+
 Ba +  MnO4 +  CN +  OH BaMnO4 +  CNO- +  H2O
 ClO3- +  H3O+ +  Br Br2 +  Cl- +  H2O
B
Match the possible colors with the given solid compounds and solutions of ions. Some colors
may be matched with more than one substance; some may not come into consideration.
Ions dissolved in water (c = 1 mol/L)
Fe2+ in acidic solution
Al3+ in acidic solution
Cu2+ in acidic solution
Cu2+ in ammoniac solution
Cl- in basic solution
Na+
Sample of solid of
Iron sulfide (FeS)
Copper sulfate (CuSO4)
Silver iodide (AgI)
Potassium sulfate (K2SO4)
Potassium permanganate (KMnO4)
Potassium chromate (K2CrO4)
Possible colors
Black, white, colorless,
yellow, yellowish brown,
light green,
blue, deep blue,
red, violet,
C
Which of the following compounds are sparely soluble?
Silver bromide, potassium nitrite, lead sulfate, calcium chloride, sodium fluoride, iron(II) sulfide.
D
Write down the formulae of the following compounds:
Barium nitrate, potassium oxalate, aluminum oxide, potassium manganate(VI), potassium aluminum sulfate, sodium carbonate decahydrate.
E
Given are sample of the elements below. Which sample of these are totally soluble in an excess
of dil. hydrochloric acid (c = 2 mol/L)?
Potassium, lead, aluminum, copper, zinc, silicium.
F
Given are sample of the elements below. Which samples of these are totally soluble in an excess
of dil. nitric acid (c = 2 mol/L)?
Potassium, lead, aluminum, copper, zinc, silicium.
Problem 3-02
Given the following galvanic cell (T = 298 K):
Cu(s) | Cu2+(aq) c = 1.00 mol/L || Ag+(aq) c = x mol/L) | Ag(s).
a)
Write down the equation for the cell reaction.
The voltage U for different values of x has been measured:
x
0.1000
0.0500
0.0100
0.0050
U in V
0.403
0.385
0.344
0.326
18
0.0010
0.285
Round 3 Test 1
b)
Plot U as a function of lg x!
c)
Write down the cell reaction for x = 0.0200 mol/L and determine the voltage.
d)
Calculate the equilibrium constant K for the cell reaction.
3.00 g of potassium iodide are dissolved in water, the solution is filled up to 50.0 cm3. 50.0 cm3 of
silver nitrate solution (c = 0.200 mol/L) is added. If this solution replaces the silver nitrate solution in
the galvanic cell the copper electrode becomes the cathode and a voltage of 0.420 V is measured.
e)
Calculate the solubility product of silver iodide.
Cu2+ + 2 e–
Ag+ + e–
Problem 3-03
Cu
Ag
E° = 0.34 V
E° = 0.80 V
Interhalogen Compounds
Compounds between different halogens are called interhalogen compounds. Besides the diatomic
compounds XY there are compounds with more than two atoms. Their formulae are generally XYn,
where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. The tendency to form
interhalogen compounds with more than two atoms rises with increasing atom mass of X and decreasing mass of Y.
a)
Give an example for an interhalogen compound with more than two atoms which should be
existent and an example which should rather not be existent.
Diatomic interhalogens can be formed from the elements. All combinations are known.
There are the following natural isotopes of the halogens in existence: 19F, 35Cl, 37Cl, 79Br, 81Br, 127 I.
b)
Write down the empirical formulae of the interhalogens XY and give the number of molecular
peaks in the mass spectrum of each compound XY.
c)
Which shape should the molecules XY3, XY5 and XY7 have following the VSEPR model?
Sketch 3-D figures.
If exposed to water interhalogens disproportionate as halogens do, too. But also without water a
disproportionation of many of these compounds may take place.
d)
Write down the equation of the reaction of XY with water (X is less electronegative than Y).
e)
Write down the equation of a possible disproportion reaction of XY in the absence of water (X is
less electronegative than Y).
Problem 3-04
Hess's Law and Reaction enthalpies
A thermally very well insulated calorimeter was filled with water of 22.55 °C. When adding 1.565 g of
zinc sulfate the temperature went up to 23.52 °C after zinc sulfate had totally dissolved.
In a second experiment the same calorimeter was filled with water of 22.15 °C. After adding and
dissolving 13.16 g of zinc sulfate heptahydrate the temperature went down to 21.84 °C.
In both cases the heat capacity of the system was 0.900 kJ/K.
19
Round 3 Test 1
a)
Calculate the enthalpy HR of the reaction
b)
Calculate the enthalpy of formation of nitrous acid (HNO2) in aqueous solution at constant pressure and temperature from the given reaction enthalpies.
(1)
(2)
(3)
(4)
(5)
NH4NO2(s)

2 H2(g) + O2(g)

N2(g) + 3 H2(g) + aq 
NH3(aq) + HNO2(aq) 
NH4NO2(s) + aq

Problem 3-05
ZnSO4 + 7 H2O  ZnSO4 · 7 H2O
N2(g) + 2 H2O(l)
2 H2O(l)
2 NH3(aq)
NH4NO2(aq)
NH4NO2(aq)
H1 = -307.4 kJ/mol
H2 = -571.7kJ/mol
H3 = -161.7 kJ/mol
H4 = -38.1 kJ/mol
H5 = +25.1 kJ/mol
Lithium
Lithium is the lightest metal and the least dense solid element. The small density is traced back to
the fact that it has as well as the other alkali metals a body centered cubic structure.
face centered cubic (fcc)
body centered
cubic (bcc)
a)
Calculate the packing fraction (in percent) of an fcc and of a bcc cell.
b)
What is the percentage difference of the density between these two structures if the atoms are
assumed to be of the same kind?
Spodumene, LiAlSi2O6, is an important source of lithium. It is chemically opened up with CaCO3 and
converted into the oxides of aluminum and lithium. When treated with water lithium hydroxide crystallizes as a monohydrate.
c)
Write down the equation of the reaction of spodumene with CaCO3.
Using hydrochloric acid lithium hydroxide can be converted into lithium chloride which is taken to
obtain pure lithium by fused-salt electrolysis. In another process lithium can be obtained by electrolysis of a solution of lithium chloride in pyridine or acetone. This solubility can be used to separate it
from sodium chloride and potassium chloride.
d)
Give the reason why lithium chloride is soluble in solvents as pyridine, acetone and alcohols in
contrast to the other alkali chlorides.
In the qualitative analysis lithium is very difficult to detect because it forms nearly no poorly soluble
compound. Quantitatively it can be determined gravimetrically as sulfate or aluminate (formal: x
Li2O · y Al2O3). In literature (H. Grothe, W. Savelsberg (1937). Über die analytische Bestimmung des
Lithiums. Z. analyt. Chem. 110, 81 – 94) the following instruction is given (translated from German):
20
Round 3 Test 1
"III. Specification of a procedure to determine lithium.
A. Precipitating agent: 50 g of potassium aluminum sulfate are dissolved in 900 mL of warm water.
The solution is cooled down and a conc. solution of 20 g of sodium hydroxide is added while
stirring and cooling, until the formed precipitate is totally dissolved. After a long time (overnight) the solution is filtered and the pH is brought to 12.6. Then the solution is filled up to 1 L.
B. Procedure of detection: The solution of lithium is brought to pH = 3. Then for each 10 mg of
lithium 40 mL of the precipitating agent is added. The solution is brought to pH = 12.6 again by
using some drops of sodium hydroxide solution (c = 1 mol/L). After a short time the solution can
be filtered. The precipitate is decanted with a lot of cold water, brought on a filter paper and
washed with cold water until the washing water does not redden phenolphthalein. The filter
paper is burned off and the residue (x Li2O · y Al2O3) weighed."
e)
What could cause the turbidity when potassium aluminum sulfate ((KAl(SO4)2 · 12 H2O) is dissolved? Why can it be dissolved again by using sodium hydroxide solution? Give the equations
for the relevant reactions.
f)
Why does the pH in this determination have be kept at exactly 12.6?
Exactly 0.1980 g of the residue of the ashing is brought into solution. The content of aluminum is
detected by complexation. 50.00 mL of a solution of Na2EDTA (c = 0.10 mol/L) is added to the solution and the Na2EDTA which has not reacted is titrated with zinc sulfate solution (c = 0.10 mol/L)
with xylenolorange (Ind) as indicator. Consumption: 15.25 mL.
g)
Write down the equation for the complexation reaction of aluminum and Na2EDTA . Use H2Y2–
for the EDTA component.
h)
Why are complexometric determinations often executed in buffer solutions (NH3/NH4Cl or
AcOOH/NaOOAc)?
i)
Arrange the complex compounds of this determination (EDTA – Zn2+, EDTA – Al3+, Ind – Zn2,+ Ind
– Al3+) in the direction of decreasing stability.
j)
In the original instruction the stoichiometric factors are replaced by x and y. Determine the empirical formula of lithium aluminate in the residue of the ashing.
Problem 3-06
The dependence of the equilibrium constant on temperature of the reaction
PCl5
PCl3 + Cl2
can be expressed by the equation
log Kp = -4374/(T/K) + 1.75·log(T/K) + 3.78.
a)
Calculate Kp at 200 °C.
The reaction proceeds under isothermal-isobaric conditions at a temperature of 200 °C and a pressure of 150 kPa in a vessel with variable volume until equilibrium is reached.
b)
Calculate p(PCl5) and p(PCl3 in equilibrium. Use in this case Kp= 0.200.
c)
Calculate the rate of conversion of PCl5 (in %).
21
Round 3 Test 1
Phosphorus pentachloride is a molecular compound only if heated above 160 °C. Below this temperature it forms an ionic solid.
d)
Which are the ions in the solid? Give the Lewis structure of these ions and draw their 3-D image.
Compounds as PCl5 and PF5 have a trigonal bipyramidal shape following the VSEPR model. In a trigonal bipyramid you can distinguish axial and equatorial positions. Nevertheless only one single fluorine signal in the 19F NMR spectrum of PF5 is detected.
e)
Account for this fact.
Problem 3-07
Isomerism
Draw the structures of all isomers with the empirical formula C3H6O and write down their names.
(Ignore stereo isomerism)
Problem 3-08
Aromatic Compounds
Given the following reaction scheme:
KMnO4
B
C
i / ii
iv / v
iv / v
i / iii
Isomerization
H2 / Pd/C
F
A
vii
CH2O
D
1. NaNO2 /
vi
2. KI
G
22
- H2O
E
Round 3 Test 1
a)
Complete the structural formulae A – G as well as the empirical formulae of the reagents i - vii.
The isomerization is known by a special name, which one?
b)
Which range of temperature has to be taken to produce G? Which product is formed if the solution (without KI) is heated to boiling?
c)
What is the resonance effect of the substituent F in case of a second substitution? Show with the
help of resonance structures in which position the substituent directs. How does your result
match with the product which forms in the reaction of F with iv/v? Explain why the product
shown is formed.
Problem 3-09
Michael Reactions
α,β-unsaturated carbonyl compounds are typical Michael systems showing an interesting reactivity.
On one hand they decolor an aqueous solution of bromine as alkenes do and, like ketones, reactions
with nucleophiles occur. On the other hand the conjugated system of a keto group and an alkene
double bond shows some unique reactions.
a)
Show the mechanism of the reaction of but-3-en-2-one with bromine. Give the (general) name of
the intermediate cation!
b)
Write down the equation for the reaction of hydrogen cyanide (HCN) with the keto group of the
butenone.
The Grignard reaction of cyclohexenone and butyl magnesium bromide gives 3-butylcyclohexane-1one in addition to the classical product 1-butyl cyclohex-2-en-1-ol:
c)
Using resonance structures account for the fact that there can be an addition not only at the
carbonyl group but also at the double bond. Is it an electrophilic or a nucleophilic addition?
Besides Grignard compounds many other organometallic reagents are used to form a C-C linkage e.g.
BuCeCl2 and Bu2CuLi both of which are produced in situ. The concept of hard and soft acids and bases (HSAB concept of Pearson) allows a good valuation whether the addition proceeds directly (at
the carbonyl carbon atom) or rather conjugated (at the β-carbon atom). The gist of this theory is that
soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and
form stronger bonds with hard bases, all other factors being equal. Cer compounds are known to be
very hard Lewis acids while copper(I) compounds are rather weak.
d)
Explain which C atom in cyclohex-2-en-1-one is harder, which one is weaker. At which C atom
would you expect a reaction with BuCeCl2 and Bu2CuLi, respectively?
23
Problems Round 3 Test 2
Third Round Test 2
Problem 3-11
Multiple Choice
With one or more correct answers even if the question is written in singular.
a)
A member of which group of compounds is generated by the oxidation of a secondary alcohol?
A
b)
D
Peroxide
E
Ketone
Cl2O2
Fe(CN)63-
B
C
VO2+
D
K2MnO4
E
H2S2O8
E
Mn5+
E
Yeast
CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.1 mol/L)
CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.05 mol/L)
CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.1 mol/L)
CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.05 mol/L)
Which ion has unpaired electrons at its disposal?
A
e)
C tert. Alcohol
Which of the following mixtures is a buffer solution?
A
B
C
D
d)
B Carboxylic acid
Which compound contains an element with the same oxidation number as chromium in
K2Cr2O7?
A
c)
Aldehyde
Cu+
As4+
B
C
Zn2+
D
Ag+
Which compound does puff up cakes?
A
CaCO3
B (NH4)2CO3
C Ca(COO)2
D NaHCO3
The image1 shows the phase diagram of water. Which of the following statements is correct?
Pressure in kPa
f)
With rising pressure
A the boiling temperature decreases and the
melting temperature increases slightly.
B the boiling temperature increases slightly and
the melting temperature decreases.
C the melting and boiling temperature increase.
D the melting and boiling temperature decrease.
E the melting and boiling temperature do not
change.
fluid
solid
gaseous
Temperature in °C
g)
Which of the following formulae represent more than one compound?
A
h)
B
C2H4
B
Image from "chemikerboard.de"
24
C2H2Cl2
C
Pt(NH3)2Cl2
D CuSO4·5H2O
E
C2H6O
D PtCl42-
E
CH4
Which of the following molecules and ions are planar?
A
1
CH4O
PH3
C
COCl2
Problems Round 3 Test 2
Problem 3-12
Buffer Action and Acidity
A buffer solution with pH = 5.8 has to be made from a diluted acid (pKS = 6.5) and its sodium salt.
a)
Calculate the ratio of amounts of acid and conjugated base.
b)
Calculate the concentration of formiate ions at pH = 4.2.
There is 1 L of a buffer solution which contains 0.1 mol of NH3 and 0.1 mol of NH4Cl. The pH value of
this solution is 9.25.
c)
Which volume of hydrochloric acid (c = 1 mol/L) and sodium hydroxide solution (c = 1 mol/L),
respectively, can be added with the result that the pH does not change more than 1?
A solution of 2.895 g of an unsubstituted weak carboxylic acid X in 500 g of water shows a freezingpoint depression of 0.147 K. If 0.957 g of sucrose (C12H22O11) is dissolved in 100 g of water a freezingpoint depression of 0.052 is detected.
d)
Which saturated carboxylic acid was used?
e)
Determine the acidity constant Ka of this acid and , the degree of protolysis.
Problem 3-13
Reactions
There are six aqueous solutions of the following compounds: NH 4Cl, BaCl2, Na2S, Pb(NO3)2,
Na2SO4 and AgNO3.
a)
Record in the table on the answer sheet whether a reaction takes place (e.g. "yellow prec." if a
yellow precipitate forms or "gas formation" ect.) or "n.r." if nor reaction takes place.
b)
Write down the equations for all reactions (indicate the aggregate state and the hydration by
using (s), (l), (g), (aq))! Is it possible to identify the original solutions only by the results of the
reactions without using more utilities? Account for your decision!
Problem 3-14
Double-Contact Process
The double contact process is used to synthesize sulfuric acid on an industrial scale. During the decisive step in this process sulfur dioxide is oxidized to sulfur trioxide:
2 SO2 + O2  2 SO3
(1).
o
o
Given are the following thermodynamic values* at 25 C and p = 1.000·105 Pa (Standard pressure).
So in J·mol-1·K-1
Cp in J·mol-1·K-1
Hf0 in kJ·mol-1
SO2 (g)
O2 (g)
SO3 (g)
-297.00
0
-396.00
249
205
257
39,9
29,4
50,7
* values taken from Atkins, Physical Chemistry 3rd Edition
Cp is the molar heat capacity at constant pressure. You can use it to determine the enthalpy of formation and the entropy of formation of a compound at a temperature differing from the standard
temperature:  Hf0(T) =  Hf0 + Cp·(T – 298 K), S(T) = S0 + Cp·ln(T/298 K).
25
Problems Round 3 Test 2
a)
Determine Kp at 600 OC by using the Cp values.
To produce SO2 at first sulfur is burnt at 1400 °C to1500 °C under oxygen deficiency conditions. Afterwards it is oxidized completely to SO2 at 700 °C in an excess of air.
b)
Why do they work at first at 1500 °C under oxygen deficiency conditions and why is the total
oxidation completed at 700 °C?
In doing so you get a mixture of 10% (V/V) of SO2, 11% (V/V) of O2 and 79% (V/V) of N2 which is practically free of SO3. This mixture is led through the contact reactor. At 600 °C and standard pressure
the equilibrium is established.
c)
Calculate the volume percentage of the components of the gas mixture at equilibrium (in %).
Determine the degree of conversion of SO2 (in %).
(Use Kp = 65.00 here. At the end of the calculation there will arise an equation of third order
with only one real solution. The result should have two decimals.
Problem 3-15
A Well Near a Volcano
The water taken from a volcano well is tested for hydrogen sulfide. A sample of 10.0 mL is taken and
all hydrogen sulfide is removed with a stream of carbon dioxide. It is absorbed in bromine water and
then the excess of bromine is removed. The acidity of the sample is titrated with a solution of NaOH
(c = 0.100 mol/L) using methyl red as indicator. 19.95 mL are being consumed.
a)
Write down balanced equations for all reactions of this method. Calculate the hydrogen sulfide
content of the water in g/L.
Another method to measure the hydrogen sulfide content works without a stream of carbon dioxide
using a different redox process. If a known amount of iodine is produced in a solution it will oxidize
hydrogen sulfide. The excess of iodine can be titrated with thiosulfate.
6.50 g of KIO3 are dissolved in water to give a solution of 1L. 10.0 mL of this solution are taken and
0.5 g of KI and 5 drops of starch solution are added. Then 10.00 mL of the volcano water are added,
too, and the mixture acidified with 10 mL of 10 % hydrochloric acid. After 5 minutes the mixture is
titrated with a solution of Na2S2O3 (c = 0.100 mol/L) until the blue colour disappears.
(Hint: Hydrogen sulfide is oxidized by bromine to form a sulfate and by iodine to form sulfur.)
b)
Write down balanced equations for all reactions of this second method. Which volume of the
Na2S2O3 solution do you expect to be consumed in the titration?
c) The silver bracelet of the technician who worked with the samples has blackened. Explain by a
reaction equation.
26
Problems Round 3 Test 2
Problem 3-16
Gaseous Compounds
A An airbag is a safety device in vehicles. It is an occupant restraint system consisting of a flexible
fabric envelope or cushion designed to inflate rapidly during an automobile collision. Older airbag
formulations contained sodium azide and other agents including potassium nitrate and silicium dioxide. An electronic controller detonates this mixture during an automobile crash forming sodium and
nitrogen. After the nitrogen inflated the cushion the temperature of the gas is appr. as low as 150 °C
due to the expansion.
a)
Write down the reaction equation of the decomposition of sodium azide.
b)
Draw the Lewis formula of the azide ion.
Since sodium metal is highly reactive the KNO3 and SiO2 react and remove it, in turn producing more
N2 gas following the (not arranged) equations (2) and (3):
Na + KNO3
 K2O + Na2O + N2(g)
(2)
K2O + Na2O + SiO2  K2SiO3 + Na2SiO3 (silicate glass) (3)
c)
Arrange the equations (2) and (3).
d)
Calculate the mass of sodium azide necessary to fill a cushion of 50.0 L (at 150 °C, 1300 hPa).
Until 1999 N2 and N3– were the only stable nitrogen containing particles which could be produced in
a larger scale. In 1999 the discovery of another such a non-cyclic particle was published: N5+.
The following structure was shown:
1,33 A
1,12A
e)
108°
166°
(K.O.Christie u.a. : N5+; ein neuartiges
homoleptisches Polystickstoff-Ion mit
hoher Energiedichte, Angew. Chem 111
(1999) 2112 – 2117)
Draw three mesomeric resonance structures of this particle which are consistent with this structure.
B 20 cm3 of a gas X are filled into a measuring tube. 80 cm3 of oxygen are added and the mixture
is ignited. When the pressure and the temperature of the beginning are restored you observe a decline of volume of 10 cm3. There is some oxygen left in the mixture after the reaction.
f)
Which of the following gases could be X?
Hydrogen, ammonia, carbon monoxide, ethene, methane.
Distinguish ϑ > 100 °C and ϑ < 100 °C.
Problem 3-17
Decay of Arsine
At 500 °C arsine, AsH3, decomposes quickly and totally in a reaction of first order to give arsenic and
hydrogen.
a)
Write down the reaction equation and the respective rate equation.
The kinetics of the decomposition was studied at a lower constant temperature in a closed tube.
27
Problems Round 3 Test 2
In the beginning of the experiment there was pure gaseous arsine in the tube with a pressure of p0 =
86.1 kPa.
After 120 minutes the pressure has increased to p120 = 112.6 kPa.
b)
Determine the pressure at the end of the decomposition.
c)
Determine the rate constant and the half-life of the reaction.
d)
What amount of time is necessary to decompose 99 % of arsine?
(Use in this case k = 1.3·10-4 s-1.)
Problem 3-18
Epoxides
Epoxides are very reactive heterocyclic compounds and therefore of great importance in the organic
synthesis.
a)
Give a reason for the great reactivity of the epoxides.
They are divided into symmetric and asymmetric epoxides:
O
O
A
B
symmetric
symmetrisch
asymmetric
asymmetrisch
Compound B can be synthesized in the following sequence of reactions starting from dimethyl sulfide:
O
S
CH3I
H
Base
C
-HI
D
O
-S(CH3)2
B
b)
Complete the structural formulae of C and D (D is a sulfur ylide, a neutral dipolar molecule containing a formally negatively charged atom attached to a heteroatom with a formal positive
charge).
c)
Show the mechanism of the reaction of D to the epoxide B using structural formulae.
Epoxides can be opened by a lot of nucleophiles to form bifunctional alkanes. The reaction of oxirane A with water, for example, catalyzed by an acid gives ethylene glycol. This reaction proceeds with
another mechanism than the reaction without an acid.
d)
Show a possible mechanism of the ring-opening catalyzed by acid using structural formulae.
e)
Complete the structural formulae of the products E – J of the following reactions:
28
Problems Round 3 Test 2
O
O
O
NH3
E
CH3OH / H+
LiAlH4
F
O
H2S
O
HBr
1. RMgX
2. H2O/H+
O
G
H
I
J
If an epoxide is treated with Lewis acids (LSkat), such as BF3 or MgI2 isomerization to a carbonyl compound takes place. Thereby a symmetric epoxide forms one isomer while asymmetric epoxides form
more of them.
f)
Draw the structural formulae of K, L and M.
Problem 3-19
O
LSkat.
K
O
LSkat.
L
M
Amino Acids
L-α-amino acids play an important role in all creatures. Histidine (on the right)
has a pKa2 value of 6.00. It is the only amino acid which contributes to the capacity of the blood buffer (pH = 7.40). Glutamic acid (2-aminopentane diacid) and
aspartic acid (2-aminobutan diacid) are important neurotransmitter. Cysteine (2amino-3-mercapto propanoic acid, mercapto = thiol) plays an important role in
the structure of proteins because it can form the stabilizing disulfide bridges.
Two cysteine molecules can be linked together with a disulfide bridge to form
dicysteine (cystine).
a)
Which species of histidine are present in the equilibrium at the pH values of 1.82 and 9.17, respectively, which are the other pKa values of histidine? Only one of the two nitrogen atoms in the
ring can be protonated, which one? Rationalize your answer.
b)
Draw L-glutamic acid in the Fischer projection.
c)
Draw 3-D structural formulae of S-aspartic acid as well as of R-cysteine. ((Hint:
the paper plane,
behind the paper plane,
in the paper plane)
in front of
In proteins and peptides the amino acids are linked with peptide bonds – the resulting compound is
in chemistry called a carboxylic acid amide. In some text books the reaction is shown in the following
way:
.
29
Problems Round 3 Test 2
This reaction proceeds actually in this way only under extreme conditions because the carboxylic
acid is not sufficiently reactive towards a nucleophilic attack (low carbonyl activity).
d)
Which electronic effect is responsible for the low carbonyl activity of the carboxylic acids?
In order to raise the reactivity of the carboxylic acids they can be converted into halides or anhydrides.
e)
Show the total mechanism of the formation of an amide between a reactive species R-COX and
an amine R‘NH2. Which geometrical structure shows the reaction center directly after the attack
of the amino group?
f)
Write down the equation for the formation of dicysteine from cysteine (with structural formulae
without stereochemical concerns). Which type of reaction is it?
Problem 3-20
Organic Puzzle
a) Draw the structural formulae of the following compounds.
A Acetaldehyde
D Methanoic acid
G Nitrile of benzoic acid
b)
B Acetone
E Aniline
H 3-Nitrobenzoic acid
C Acetophenone
F Benzaldehyde
I Phenol
Assign each compound (using the letters of the table above) to a cell of a 3x3 table given on the
answer sheet.
Consider thereby the criteria below.
1. Line:
The aqueous solution shows an acidic reaction.
2. Line:
The compound can react with itself in an aldol reaction.
3. Line:
It in a monosubstituted aromatic compound.
1. Column: The compound contains an aldehyde group as a functional group.
2. Column: It is an aromatic compound, the substituent(s) of which directs newly incoming substituents into meta position.
3. Column: Compounds without such criteria.
30
Problems Round 4 (theoretical)
Fourth Round (theoretical problems)
(A periodic table and the same list of formulae and data as in the third round were provided)
The top 15 of the 3rd round are the participants of the 4th round, a one-week practical training.
Problem 4-01
Traces of Water
The compound KPbI3 can be used to detect minimal amounts of water qualitatively. To detect water
quantitatively other methods are used.
Method A: Heating / Annealing
The substance is heated until its mass is constant.
a) Record preconditions of the substance when using this method.
Method B: Gaseous (reaction) water
The enclosed water is bound chemically or physically. The increase of mass of the absorbing agent
such as sulfuric acid and anhydrous calcium chloride is determined. This method is also used to dry
gasses.
b)
Give the reason why hydrogen sulfide, hydrogen iodide and ammonia should not be dried by
sulfuric acid. Write down respective reaction equations.
c)
Give the reason why ammonia may not be dried by calcium chloride.
Method C: Calcium carbide (CaC2)
The water containing sample is brought to reaction with calcium carbide and the reaction product is
led into an ammoniac copper(I) chloride solution. The red precipitate is filtered off, dried to mass
constancy and weighed.
d)
Draw the Lewis structure of the carbide anion. Give the formulae of three isoelectronic species.
e)
How does copper(I) chloride exist in an ammoniac solution?
f)
Write down the equation of all reactions of this method.
Method D: Iodometric (Karl Fischer method)
In 1935 Karl Fischer published a method to determine water based on the reaction between iodine,
sulfur dioxide and water previously published by Bunsen.
g)
Give the equation of the "Bunsen" reaction!
In the Fischer method the water containing sample is brought to reaction with methanol, pyridine,
sulfur dioxide and iodine following formally the equation
H2O + SO2 + 3 C5H5N + I2 + CH3OH
C5H5NHCH3OSO3 + 2 C5H5NHI .
The endpoint of the titration is reached when a permanent brown colour occurs.
h)
What is responsible for the brown colour? Why can the iodine-starch reaction not to be used?
i)
Which function has the pyridine?
31
Problems Round 4 (theoretical)
To determine the water content the sample is added to a solution of iodine and sulfur dioxide in
water free methanol which contains pyridine and then titrated with a solution of iodine in alcohol.
As the change of colour at the endpoint (bright yellow  brown) is difficult to detect visually the
change is nowadays detected coulometrically. During the titration iodine is produced in an electrochemical cell until no more iodide is formed. The water content can be calculated by the used
amount of charge.
A
B
Sample 1
1.65 mL
1.45 mL
Sample 2
1.62 mL
1.43 mL
Sample 3
1.60 mL
1.45 mL
j)
The titer of the Karl-Fischer solution is given in water equivalents in mg/mL and amounts to t = 4.8 mg/mL. Samples of 10 g
of two different food oils are investigated. The results are given
in the table on the left.
Calculate the mass percentage of water in the oils.
In another oil C a water content of 0.09 % is found in a coulometric titration.
k)
Which amount of charge (in coulomb) is consumed in an investigation of 10 g of oil C?
Problem 4-02
Electrochemistry
A Latimer diagrams are plots of reduction potentials of half reactions including each of the different oxidation states of one element. Normally the species with the highest oxidation state is placed
on the left, going to the right the oxidation state decreases. The different states are connected with
arrows on which the reduction potential of the half reaction is written. These may refer to standard
conditions (25 °C, pH = 0, c = 1 mol·L−1) or to any other condition (e.g. pH = 14).
Example:
a)
0.926 V
1.154 V
[AuCl4]–  [AuCl2]–  Au
(Standard conditions)
Calculate the standard potential x = E°([AuCl4]–/ Au).
Gold does not react with nitric acid but does with aqua regia, a 3:1 mixture of conc. hydrochloric acid
and conc. nitric acid, which was developed by alchemists to "dissolve" gold.
In this reaction with aqua regia the complex [AuCl4]– is formed.
b)
Calculate the complex formation constant of [AuCl4]–, Kco =
c([AuCl4 ]− )/c0
3+
(c(Au )/c0 ) ·(c(Cl− )/c0 )4
using the
result of a) and E°(Au3+/Au) = 1.50 V.
In an acidic solution (pH = 0) the following standard potentials are existent:
ClO4–/ ClO3–
ClO3–/ ClO2–
ClO2–/ HClO
HClO/ Cl2
E° in V
1.20
1.18
1.65
1.63
c)
32
Cl2/ Cl–
1.36
Draw the Latimer diagram.
Check whether ClO3– disproportionates under these conditions to form ClO4– and Cl–. Write down
the equation for a disproportionation reaction if the case may be.
Problems Round 4 (theoretical)
B
d)
In which span of pH-values can hydrogen peroxide oxidize Cer3+ions?
E°(Ce4+/Ce3+) = 1.61 V
E°(H2O2, H+/H2O) = 1.76 V
On the other hand hydrogen peroxide can be oxidized by strong oxidation reagents such as potassium permanganate in acidic solution.
e)
Write down the equation of this reaction.
Problem 4-03
Metals and Electrons
When atomic orbitals are filled up with electrons the lowest available orbital is fed first and then the
following in the direction of rising energies. Thereby Hund's rule as well as the Pauli principle has to
be considered. There are exceptions from these rules.
a)
Give two examples for elements with electron configurations (in the ground state) other than
the expected regular one. Write down the expected and the observed configuration. (Use the
abbreviation for full electron shells such as [He] for 1s2 or [Ne] for s22s22p6 etc.)
The reason for this deviation is the stability of certain electron configurations.
b)
Which electron configurations are energetically especially favoured?
These favoured electron configurations play a decisive role in the electron configuration of metal
cations.
c)
Write down the electron configurations of the following metal cations.
(Use the abbreviation for full electron shells as mentioned above.)
i) Fe3+ ii) Mn3+ iii) Pd4+ iv) Cr3+
v) Fe2+
vi) Pb2+ vii) Au3+ viii) Co2+ ix) Cu+
x) Ti2+
Discrete metal ions as given in part c) exist only formally. In reality e.g. in solids or complex compounds they always have a coordination sphere and are surrounded by other particles which form a
regular coordination polyhedron. You often find the coordination numbers four and six.
d)
Draw 3-D figures of the polyhedrons for the coordination numbers four and six. (Don't consider a
hexagon and a three-sided prism)
e)
Draw the theoretically possible stereoisomers of the complex compounds MX2Y2, MX4Y2 and
MX3Y3 and give their names (M: central particle; X and Y: monodentade ligands).
All degenerate systems try to reduce orbital degeneracy. The degenerated d-levels of an isolated
metal cation split as soon as they come under the influence of a ligand field into levels of higher and
lower energy.
f)
Describe the splitting off of the d-electron energy levels in an octahedral ligand field. What are
the effects on the different d-orbitals? Draw an energy scheme which shows this issue as accurately as possible.
g)
With which d-electron configurations high- and low-spin configurations occur at all?
h)
Give the number of unpaired electrons which exist in a high-spin and in a low-spin state for the
cations of part c) in an octahedral ligand field.
33
Problems Round 4 (theoretical)
Problem 4-04
Calculations around Kinetic and Energetic
In a gas flow substance A exists with a partial pressure of p(A) = 8.9·10-4 bar. A is in equivalence with
A2 (2 A
A2) which is gaseous, too, with Kp = 2.1·103.
a)
Determine the ratio p(A)/p(A2).
keff
Given the reaction A + B 
C + D with the rate constant keff.
The following reaction sequence is assumed to be the reaction mechanism
A + B
k1
k-1
k2 C + D .
AB 
The reaction rates v1 and v-1 are approximately of the same size. Additional is k1/k-1 = 15 (mol/L)-1
and k2 = 25 s-1.
b) Calculate keff using the assumption of steady state equilibrium.
c) Which is the necessary precondition to use the steady state approximation?
The hydrolysis of urea follows the equation
(NH2)2CO(aq) + H2O(l)  2 NH3(aq) + CO2(aq) .
d)
Calculate the equilibrium constant K for this reaction at 298 K.
H° in kJ/mol
-1 -1
S° in Jmol K
e)
(NH2)2CO(aq)
-317.7
176
H2O(l)
-286.0
68
NH3(aq)
-80.9
110
CO2(aq)
-413.1
121
Calculate G if the following concentrations exist at 298 K
c((NH2)2CO(aq)) = 0,85 mol/L
c(CO2(aq)) = 0,097 mol/L
c(NH3(aq))= 0,02 mol/L
k3
In the equilibrium D + E
F the reaction as well as the back reaction are elementary reack-3
tions. c(D), c(E) and c(F) are time dependent concentrations in the process of the spontaneous reaction of D and E.
f) Determine the dependency of G on v3 and v-3 which are the rates of the reaction and the back
reaction.
k3
At a certain time in the course of the reaction D + E
k-3
rate is defined as vob = v3- v-3 .
At a certain time let be
vob/ v3 = 0.5
c(D) = 0.4 mol/L
c(E) = 0.9 mol/L
g)
F from part e) the observed reaction
c(F) = 1.8 mol/L.
Calculate the equilibrium constant K for this reaction at 298 K.
Problem 4-05
Three Compounds
Given are the compounds AB3, CA and CB2. The mass percentage of A in AB3 is 23.81 %, that of B in
CB2 73.14 %.
a) Calculate the mass percentage of C in CA.
34
Problems Round 4 (theoretical)
b)
Determine the elements A, B and C. Write down the formulae and names of the given compounds.
Problem 4-06
Complexes and More
A
Insoluble Prussian blue is a coordination compound. The crystal structure and the empirical formula
can be derived from one octant (on the left) of the cubic unit cell (drawn with iron ions only).
(on the edges)
(on the edges)
(H2O in the middle of the octant)
Indicate in all of the questions a) to d) the oxidation states of the iron ions as FeII, FeIII.
a) Determine the empirical formula of Prussian blue using the octant above. Show shortly how you
found your result.
Insoluble Prussian blue forms only at high concentrations of iron ions. At lower concentrations socalled soluble Prussian blue K[FeIIIFeII(CN)6] is formed.
It originates from the combination of an iron(II) solution with a solution of potassium hexacyanoferrate(III) as well as from an iron(III) solution with a solution of potassium hexacyanoferrate(II).
b)
Write down the equation for these two reactions.
c)
Taking these reactants find an equilibrium reaction which explains the formation of the identical
product.
Comparable with the reaction of part b) there is a reaction between a solution of iron (II) with potassium hexacyanoferrate(II) which leads to a white precipitate.
d)
Write down an equation of the reaction which leads to the white precipitate. Explain why this
compound is not colored in contrary to soluble Prussian blue.
The orientation of the cyano groups in the compound of part a) (see fig. above) can be derived with
the help of the HSAB concept. It says that the combinations of weak with weak and of hard with
hard acids and bases lead to more stable adducts than mixed combinations. Iron(III) is classified as
hard, iron(II) as weak. The hardness of bases decreases within the row
F > O > >N, Cl > Br, H > S, C > I, Se > P, Te > As > Sb.
35
Problems Round 4 (theoretical)
e)
Fill the cyano groups in their expected orientation following the HSAB concept into the empty
circles on the answer sheet (C for carbon, N for nitrogen). Determine the coordination spheres of
iron(II) und iron(III) in the compound.
B
A very diluted solution of copper(II) chloride in water has a light blue color. By adding hydrochloric
acid the solution turns green and with rising concentration of hydrochloric acid intensively greenbrown.
f)
Which species of copper(II) chloride exist predominantly in the diluted aqueous solution and in
the solutions in half-concentrated and concentrated hydrochloric acid?
In analytical chemistry copper is detected in the form of a deep blue colored ammine complex. But it
can be recognized with a borax bead as turquoise metaborate (Cu(BO2)2), too. In literature you find
different formulae for borax with the empirical formula B4H20Na2O17:
As decahydrate Na2B4O7 · 10 H2O,
octahydrate Na2[B4O5(OH)4] · 8 H2O
and as mixed oxide decahydrate Na2O · 2 B2O3 · 10 H2O.
g)
Show with the help of a Lewis structure why the octahydrate is the most reasonable. Attach
charges to the atoms if necessary.
In the separation scheme of cations copper(II) ions are precipitated with hydrogen sulfide as black
copper sulfide. However, this compound is not copper(II) sulfide but a mixed compound of copper(I)
and copper(II) ions which contains another sulfur containing anion besides sulfide anions (S2–).
h)
Which other anion does the compound contain besides the sulfide anion? Give a more exact
formula than CuS.
In the Deacon process to produce chlorine by oxidizing hydrogen chloride using oxygen from the air
copper(II) chloride is used as catalyzer.
i)
Write down the equation for the total reaction of the Deacon process.
j)
Give reaction equations which show the catalytic effect as well as the regeneration of copper(II)
chloride.
Problem 4-07
Many Questions Concerning Thermodynamics
2 mol of oxygen at 273 K have a volume of 11.2 dm3. The gas can be regarded as perfect with the
heat capacity of CV = 21.1 Jmol-1K-1 which is supposed to be independent of temperature.
a)
Calculate the pressure of the gas.
b)
Give the different meanings of Cp and CV. Explain why these values have to be different.
Calculate Cp.
The sample of gas mentioned above is heated reversibly to 373 K at constant volume.
c)
36
How much work is done to the system?
Problems Round 4 (theoretical)
d)
Calculate the rise of the internal energy.
e)
Calculate the heat which was added to the system.
f)
What is the final pressure?
g)
What is the increase in enthalpy H?
The gas sample (2 mol at 373 K in 11.2 dm3) is allowed to expand maintaining the temperature
against a piston that supports a pressure of 2 atm.
h)
Calculate the work done by the expansion.
i)
What is the change in internal energy and in enthalpy of the gas?
j)
Calculate the heat absorbed by the gas.
The dependence of the boiling point of methane on pressure is well described by the empirical equation:
k)
log(p/bar) = 3.99 –
443
Ts /K - 0,49
Determine the boiling point of methane at a pressure of 3 bar.
The difference in internal energy of liquid and gaseous methane at the boiling point of 112 K at atmospheric pressure is 7.25 kJ/mol.
An object is cooled by the evaporation of CH4(l).
l)
What volume of CH4(g) at 1.000 atm must be formed by the liquid to remove 32.5 kJ of heat
from the object?
Problem 4-08 Synthesis of (–)-Muscone
(–)-Muscone is the primary contributor to the odor of musk. It is an oily liquid with a characteristic
smell. Natural muscone was originally obtained from musk, a glandular secretion of musk deer,
which has been used in perfumery and medicine for thousands of years. Since obtaining natural
musk requires killing the endangered animal, nearly all muscone used in perfumery today is synthetic.
One synthesis of (−)-muscone begins with commercially available (+)-citronellal, a monoterpene
aldehyde, which is mainly found in citrus fruits. The scheme on the next page shows the way of synthesis.
Hints:
- In this scheme compound A reacts exclusively to compound B and compound C exclusively to D
aas well as compound G exclusively to (–)-muscone.
- Compound V is a side product.
- Compound G exists as a mixture of two isomers.
- (–)-Muscone has the empirical formula C16H30O.
37
Problems Round 4 (theoretical)
10-bromo-dec-1-ene
then r.t.
then r.t.
Grupps cat. I
(-)-Muscone
r.t. = room temperature
-
TBDMSCl
Protective reagent for hydroxy groups
Si
Cl
-
TBAF
splits up silyl ether
N+ F-
-
THF
-
Grupps cat. I
Solvent
O
Cl
Cl
P(Cy)3
Ru
P(Cy)3 Ph
catalyzes the metathesis of olefins, e.g.:
R1
R1
R2
R2
Cy = Cyclohexyl-Rest
38
+
R3
R3
R4
R4
Kat.
Cat.
R1
R2
R3
R4
2
Problems Round 4 (theoretical)
a)
Determine and draw the structural formulae of the compounds A to G as well as the structural
formula of (-)-muscone.
During the reaction of (+)-citronellal to compound A in a so-called carbonyl-ene-cyclisation compound V (isopulegol) forms as a side product. Under the terms given in the scheme above the Lewis
acid X is formed which catalyzes the carbonyl-ene-cyclisation (the stereoselectivity of this reaction
stays unaccounted):
O
H
X
(+)-Citronellal
OH
V, Isopulegol
b)
Give the empirical formula of X and write down the reaction equation of the formation of X.
c)
Map the mechanism of the carbonyl-ene-cyclisation taking the catalyzing effect of X into consideration.
The formation of F succeeds with a (deep) red chromium(VI) species Y which is formed in situ. It
oxidizes E to compound F and is itself reduced to a green compound.
Y is formally the anhydrate of the chromium acid.
d)
Write down the equations for the reduction and the oxidation as well as the complete equation
for the redox reaction between the compounds E and Y. Use the abbreviations R and R' for the
substituents outside the reactive center.
Step F to G is a so-called cyclisation metathesis. A gaseous compound Z forms besides compound G.
e)
Give the name of compound Z?
Problem 4-09
Small but Powerful
Carbocyclic compounds are often found as basise of biologically active reagents and functional materials. Many ways are known to synthesize the especially stable five- and six-membered rings but
the preparation of larger and smaller rings is a synthetic challenge. The reason for the difficulties
with smaller rings is their low stability and thus their high reactivity.
The bonding situation in a cyclopropane ring is described by bent bonds. The overlapping orbitals
between two carbon atoms can't point directly to each other (as they do in alkanes); rather, they
overlap in an angle. The result is a banana-shaped bond shown on the next page.
a)
Compare the bond angel and the bond length of cyclopropane and the more stable cyclohexane.
Draw a cyclopropane and a cyclohexane ring (in chair conformation with hydrogen atoms) in a
3-D plot. Account for the different stability of the C-C bonds in cyclopropane and cyclohexane.
39
Problems Round 4 (theoretical)
Banana-shaped bond in cyclopropane
Because of the higher π-character of the C-C bonds in cyclopropane rings addition reactions may
proceed analogue to those of alkanes. The electrophilic hydrobromation is an example.
b)
Write down the equation of the reaction of methyl cyclopropane with hydrogen bromide.
Give the mechanism of the hydrobromation and use it to explain which product is formed preferentially.
What is the name of the rule of this regioselective formation?
c)
Draw an energy profile of this reaction (ΔRH° < 0, energy depending on the reaction coordinate)
and attach reactants, products, transition states and intermediates (if existent).
Generally cyclopropane rings are generated via cycloaddition of carbenes and carbenoids to olefins.
Carbenoids are substances similar to carbenes and show a comparable reactivity.
d)
Give a common form of the Lewis structure of a carbene (with substitutes = R). What is the hybridization state of the carbon atom in the center of the carbene?
The best method for preparing cyclopropanes is by a process called the Simmons-Smith synthesis:
An addition of CH2-Zn-I to a C=C double bond. In this reaction the new C-C bonds form in cis position
of the original double bond.
e)
Draw the structural formula and give the name of the product of the Simmons-Smith reaction of
(Z)-1,2-diphenylethene.
Another compound to synthesize cyclopropane rings starting from olefins is diazomethane (CH2N2),
which is also used for the preparation of the hydrocarbon 4 (C13H20). 4 has a tetrahedral shape. The
1
H- and 13C-NMR spectra show two resp. three signals. The scheme of this reaction is shown below:
.
Ph3P Br2/DCM
CO2Et
LiAlH4/Et2O
1
-15°
99%
t
BuOK/DMSO
20°C, 6h
3
2
95%
CH2N2 (10 eq),
Pd(OAc)2
-20° 25°C
sechsfach
wiederholt
repeated six
times
60%
DCM = dichloromethane, DMSO = dimethyl sulfoxide
40
20°C, 6h
4
92%
Problems Round 4 (theoretical)
f)
Give the structural formulae of the compounds 1 to 4.
g)
Mark in a figure of 4 those C and H atoms which are responsible for the NMR signals.
Problem 4-10
Carbonyl Compounds as Reactants and Products
Carbonyl compounds are valuable resources for organic synthesis because they react with many
different compounds, often selectively. In the following scheme some reactions starting with a butyl
alcohol (1) are depicted:
iPr: Isopropyl
cHex: Cyclohexyl
DMP:
m-CPBA:
a)
Draw the structural formulae A to G.
b)
Give a possible method for i) to form B directly from butyl alcohol.
Compound F is an epoxide.
c)
Which reagents ii) are needed to synthesize the diol 2 and then to convert the diol 2 with iii) to
A? Give the name of the reaction of diol 2 to A.
(Help: Compound iii) contains a halogen atom in the highest oxidation state.)
d)
Which product do you expect if compound C is brought to reaction with MeLi?
Draw the structural formula.
e)
Why is it not possible to get compound D by bringing B directly into reaction with MeLi?
Finally compound D is converted to 5.
f)
Propose the reaction conditions iv) for the reaction of D to 5.
g)
Show the mechanism for this reaction and give the names of the class of substances of the intermediate and the product.
41
Problems Round 4 (practical)
Fourth Round (practical problems)
Problem 4-11 Gravimetric Determination of Zinc as Zn(NH4)PO4
Equipment:
2 x 400 mL beaker, 25 mL pipette with pipette control, 50 mL graduated cylinder, 100 mL narrownecked bottle, Bunsen burner with tripod and tile, glass rod, 2 glass filter crucibles, suction flask
with rubber ring, pressure tubing, vacuum attachment, desiccator with drying agent, precision balance, pen
Chemicals:
Test solution (100 mL volumetric flask)
Solution of ammonia, c(NH3) = 2 mol/L
Ammonium chloride, NH4Cl(s),
diluted hydrochloric acid, c(HCl) = 2 mol/L
Solution of diammonium hydrogen phosphate, w((NH4)2HPO4) = 10 %
Indicator solution of methyl red in Ethanol, w(C15H15N3O2) = 0,05 %
Demineralized water
Procedure:
The test solution has to be filled up with demineralized water to the calibration mark and mixed
well. 25 mL of this solution are pipetted into a 400 mL beaker. Approximately 150 mL of demineralized water are added.
Then 25 mL of diluted hydrochloric acid, 2-3 full spatulas of ammonium chloride, 25 mL of a solution
of diammonium hydrogen phosphate and some drops of methyl-red indicator solution are added.
Heat to boiling and add dropwise diluted ammonia until a colour change to yellow-orange can be
observed. Then stir with the glass rod until the precipitate has formed crystals or is well sedimented.
While the solution cools down to room temperature the glass filter crucibles are marked with the
pen, weighed and the values are listed.
The precipitate is sucked through a glass filter crucible, washed with a small portion of cold demineralized water and then dried at 130°C for one hour in a drying oven.
The crucible is cooled down to room temperature for about 20 to 30 minutes in an exsiccator and
then weighed again.
a)
Write down the number of your sample on the answer sheet.
b)
Write down the results of your weighing.
42
Problems Round 4 (practical)
c)
Calculate the mass concentration of zinc in mg/L in your test solution.
d)
What kind of compound is zinc ammonium phosphate, a mixed crystal, a mixture of crystals, a
double salt or an alloy? Account for your decision.
e)
What happens if you anneal the precipitate? Write down the reaction equation.
Problem 4-12
Standardization of a Na2EDTA Solution
This solution will be used in problem 4-13.
Equipment:
250mL volumetric flask with a solution of Na2EDTA (concentration unknown), stopper, 20 mL pipette
with pipette control, 2 Erlenmeyer flasks (wide mouth), spatula, 25 mL burette with funnel and
clamp, stand.
Chemicals:
Solution of Na2EDTA, (c(Na2EDTA) ≈ 0,1 mol/L when filled up)
Solution of ammonia, w(NH3) = 25 % (in the hood)
Indicator buffer pills
Standardized solution of zinc sulfate, c(ZnSO4) = 0.097 mol/L
Demineralized water
Procedure:
The flask with the solution of Na2EDTA of unknown concentration has to be filled up with demineralized water up to 250 mL. Mix thoroughly. Transfer exactly 20 mL of this solution into an Erlenmeyer
flask and fill up to 100 mL. Add an indicator buffer pill and - after it is dissolved - 2 mL of the ammonia solution (w(NH3) = 25 %). Titrate speedily with the standardized solution of zinc sulfate (c(ZnSO4)
= 0,097 mol/L). The end-point is given by the color change from green to red.
Problems:
a)
Write down the label code of your volumetric flask on the answer sheet.
b)
Note the consumption of the standardized solution of zinc sulfate.
c)
Calculate the concentration of your Na2EDTA solution.
43
Problems Round 4 (practical)
Problem 4-13
Complexometric Determination of Nickel
Equipment and Glassware:
Volumetric flask (100 mL) with stopper with test solution, volumetric pipette (20 mL), pipettes control, 1 burette (25 mL) with funnel, 1 measuring cylinder (50 mL), stand and clamp, spatula, 2 conical
(Erlenmeyer) beakers (300 mL, wide mouth)
Chemicals:
Test solution containing nickel a volumetric flask
Dil. solution of ammonia, c(NH3) = 2 mol/L
Trituration of murexide indicator
Solution of Na2EDTA·2 H2O, c(Na2EDTA)  0.1 mol/L (from problem 4-12)
Demineralized water
Procedure:
The flask with the test solution has to be filled up to 100 mL. The solution has to be mixed well.
20 mL of this solution are transferred with a pipette to a conical beaker (300 mL, wide mouth) and
15 mL of the solution of ammonia are added.
Drops of the indicator solution are added until intense yellow color occurs (approximately 8 drops).
If the solution in orange the pH value is not high enough and additional solution of ammonia has to
be added.
The mixture is filled up with demineralized water to about 100 mL and then titrated with the solution of Na2EDTA.
There will be a sharp change of colors from yellow to violet. This color has to persist.
Problems:
a)
Write down the label code of your volumetric flask on the answer sheet.
b)
Record the consumption of the standard solution of Na2EDTA.
c)
Calculate the mass concentration  of nickel in your tested solution (in mg/L).
Problem 4-14
Qualitative Analysis
You find the following mixtures of salts in seven beakers:
BaCl2/NaCl - AgNO3/Cu(NO3)2 – FeCl3/CuCl2 – KSCN/KI – KIO3/K2SO4 – Na2CO3/NaOAc –
(NH4)2SO4/FeSO4.
Equipment:
25 test tubes, test tube holder, 3 Pasteur pipettes, pH paper, spatula
Chemicals:
Seven beakers with mixtures of salts (labelled from A to G)
44
Problems Round 4 (practical)
Dil. solution of ammonia, c(NH3) = 2 mol/L
Dil. nitric acid, c(HNO3) = 2 mol/L
Demineralized water
Procedure:
Dissolve each mixture with about 30 mL of demineralized water. Combine at any time two of the
solutions in a test tube and note the observation. Repeat this procedure with all possible combinations. You may use the dil. solutions of ammonia and nitric acid as aids as well as the pH paper.
Problems:
a)
Write down the label code of your test mixture.
b)
Report your observations on the answer sheet.
c)
Assign the salt mixtures to the beakers.
The following reduction potentials are given (pH = 0):
2+
+
Cu / Cu
Cu2+ / Cu
I2 / I–
Fe3+ / Fe2+
E° in V
0.16
0.34
0.54
0.77
+
Ag / Ag
HIO3 / HIO
Cl2 / Cl
E° in V
0.80
1.13
1.34
Take possible acid/base reactions into consideration.
Additional safety precautions:
The mixtures contain toxic heavy metals, such as barium.
Do not pour the solutions into the sink.
45
Answers
Part 2
The answers to the problems of the four rounds
The solutions are more detailed than expected from the pupils. That may facilitate their
comprehension in case of mistakes. Furthermore future participants should use this booklet
to become acquainted with the problems of the competition and their solutions.
46
Answers Round 1
Answers Round 1
Solution to problem 1-01
a)
b)
Natural isotopes: 126𝐶 , 136𝐶 und 146𝐶 . They differ in the number of neutrons in the nucleus.
14
14
1
C is produced from nitrogen by cosmic radiation in the atmosphere:
7N + 0n →
c)
14
6C
d)
𝑁=
N = number of nuclei at time t, N0 = number of nuclei at the begin (t = 0), λ = decay constant
The half-life is the time it takes for the number of nuclei to fall to half its initial value. It is constant for a
nuclide and characteristic for it.
e)
14
7N +
𝑁0 · 𝑒 –𝑡
→
14
6C
+ 11H
e–
N0
The formula for half-life can be derived by setting N = N0/2:
2
= N0 · e–t1/2

t½ =
ln 2

=
0,693

f)
g)
Never, N = 0 will never be reached.
12
13
14
Carbon has two stable, nonradioactive isotopes: C and C, and one radioactive isotope, C. Radiocar14
bon dating is a radiometric dating technique that uses the decay C to estimate the age of organic mate14
rials. The concentration of C in the atmosphere might be expected to reduce over thousands of years.
14
However, C is constantly being produced in the lower stratosphere and upper troposphere by cosmic
rays. Thus the proportion of radioactive to non-radioactive carbon in the atmosphere is constant and in
living organic material too. Once an organism dies the natural carbon exchange is not continued and the
14
14
ratio of C in the material decreases. The comparison of the actual content of C in some organic material with its content in the atmospheric carbon leads to the age of the organic material. In doing this you
may compare the number of decays of 1 g of carbon per minute.
h)
The decay rate is dependent on the number of atoms.
N = N0 · e–t 
and with  =
i)
ln 2
t = ln
N0
N
t=
:
5730
·
1

5730
ln2
years · ln
15
14,48
= 292 years
2013 – 292 = 1721
The treasury map originates from the year 1721, it could be authentic.
14
After 65 m years the ration of C is so small that hardly a decay can be detected. Furthermore it cannot
be assumed that the ratio of carbon isotopes has been stable in such a long period.
Remark: Nevertheless the age of the bones of dinosaurs can be estimated by the decay of radioactive
238
87
40
isotopes of other elements with a longer half-life such as U, Rb und K.
Solution to problem 1-2
a)
The "characteristic composition of fatty acids of a fat" shows the mean distribution of fatty acids in a fat.
For example, a fat could consist of the following three esters
O
O
O
O
O CH2
O CH
O CH2
O
O
O
O CH2
O
O CH
O
O CH2
O CH2
O CH
O CH2
3 x Oleic acid
2 x Oleic acid 1 x Stearic acid
1 x Oleic acid, 2 x Stearic acid
Then the characteristic composition of fatty acids is 66.7 % of oleic acid and 33.3 % of stearic acid.
b)
Capric acid
10 : 0
COOH
C10H20O2
Linoleic acid
C18H32O2
COOH
18 : 2
47
Answers Round 1
Linolenic acid
C18H30O2
Erucic acid
C22H42O2
c)
18 : 3
COOH
22 : 1
COOH
The residues of the fat molecules experience van der Waals forces. Because van der Waals forces operate
only at short distances they are strongest in molecules which chains can pack together closely. Saturated
residues can arrange linearly which leads to strong intermolecular forces. These fats are solid.
Residues with double bonds are mostly cis configurated. Thus a linear arrangement is not possible and
the interactions are lower. These fats are rather liquid.
d)
e)
O
1
R
O
R2
R3
O
O
CH2
R
+ NaOH
O
O
CH
1
O
CH2
O
R2
O
O–
R3
O
– Na+
O
O
CH2
R
CH
R2
CH2
R3
OH
–
1
O
O
O
CH2
O
CH
–
OH
–
OH
O
CH2
OH
O
1
R
+ NaOH
R2
O
O
CH2
O
+ H2O
CH
O
– H2O
–
R
3
O
O
O– Na+
O
CH2
– OH–
R3
R1
O– Na+
+
R2
O
O
CH2
O
CH
HO
H
O
CH2
H
R1, R2, R3 = any alkyl- or alkenyl residue of a fatty acid
f)
g)
The deprotonation of a carboxylic acid by hydroxide ions is almost irreversible, thus the reaction runs towards the carboxylates.
The attack of the hydroxide is hindered by the long chain ester.
A tenside has two different ends, a hydophobic and a hydrophilic end. So they have the property to accumulate at the boundary surface of liquids and thus decrease the surface tension.
When tensides are dispersed in water the long hydrocarbon tails cluster together on the inside of a tangled, hydrophobic ball, while the ionic heads on the surface of the cluster stick out into the water layer.
These mostly spherical clusters are called micelles. If there is a higher concentration cylindrical micelles
and block micelles may be formed.
Tenside molecule
48
Answers Round 1
h)
Spherical micelle
Cylindrical micelle
Block micelle
Grease and oil droplets are solubilized in water when they are coated by the nonpolar tails of the tenside
in the center of micelles. Once solubilized, the grease and oil can be rinsed away.
Soaps are anionic tensides. There are four kinds of tensides.
Anionic tensides: a long hydrophobic hydrocarbon tail is connected
–
with a hydrophilic negatively charged group.
Cationic tensides: a long hydrophobic hydrocarbon tail is connected
+
with a hydrophilic positively charged group.
Ampholytic tensides: a long hydrophobic hydrocarbon tail is connected
+/–
with an ampholytic group (positively and negatively charged).
Nonionic tensides: a hydrophobic residue is connected with an uncharged group of polyglycol ethers.
Anionic tensides
Soaps
Alkylbenzene sulfonates (ABS, LAS)
R CH2 COONa
1
R
SO3Na
CH
R2
R1
CH SO3Na
R2
Alkane sulfonates (AS)
-Olefin sulfonates (AOS)
Ester sulfonates (ES)
Fatty alcohol sulfate (FAS)
Fatty alcohol ether sulfate (FAES)
R CH
CH
CH2 SO3Na
R CH COOMe
SO3Na
R CH2
R O
O
SO3Na
CH2 CH2 O
n
SO3Na
Cationic Tensides
H37C18
Distearyl dimethylammonium chloride (DSDMAC)
CH3
Cl–
N+
H37C18
H3C
Dodecyl dimethylbenzyl ammonium chloride
CH3
CH3
N+
H25C12
Cl–
CH2
O
Esterquats (EQ)
CH2 CH2 O C R
+
H3C N CH2 CH2 OH
Cl–
CH2 CH2 O C R
O
Amphoteric Tensides
CH3
Betaines
R
N+
CH2
COO–
CH3
R = C12 – C18
49
Answers Round 1
CH3
Sulfobetaines
N+
R
CH2
CH2
SO3–
CH3
R = C12 – C18
Nonionic Tensides
Fatty alcohol polyglycol ether (FAE)
Alkylphenol polyglycol ether (APE)
R O (CH2 CH2O)n H
R
O
(CH2 CH2O)n H
Fatty alcohol polyethylenglycol polypropylenglycol ether
H
R
(CH2 CH2O)n
(CH2 CO)m
H
CH3
O
Fatty acid ethanolamide
CH2
CH2
OH
CH2
CH2
OH
R C N
Solution to problem 1-3
a)
Air, oxides, carbonates, silicates, sea water, water, biosphere or other compounds (or individual compounds)
b)
Electrolysis of water:
2 H2O
2 H2 + O2
Thermic decomposition of peroxides:
2 BaO2
Catalytic decomposition of hydrogen peroxide (Pt,
MnO2):
2 H2O2
2 H2O + O2
Catalytic decomposition of oxygen containing compounds:
2 Ag2O
2 Au2O3
160°C
4 KClO3
KClO4
300°C
4 Ag + O2
4 Au + 3 O2
700 °C
500 °C
2 BaO + O2
 > 160°C
>
3 KClO4 + KCl  > 400°C
KCl + 2 O2
>
the decomposition of KClO3 proceeds already at
150 °C with MnO2 as catalyst:
KClO3
KClO3 + 3 MnO2
3 MnO3
KCl + 1,5 O2
KCl + 3 MnO3
3 MnO2 + 1,5 O2
c) Nitrogen is diamagnetic, oxygen paramagnetic. The reason is the existence of unpaired electrons in
the oxygen
molecule.
50
Answers Round 1
Energy
Energy
Oxygen
Nitrogen
d)
Triplet oxygen
2 unpaired electrons with equal spin
½+½=1
2·1+1=3
Total spin S
Multiplicity M
Singlet oxygen
2 unpaired electrons with different spin
–½ + ½ = 0
2·0+1=1
The names follow the multiplicity: M = 1 Singlet, M = 2 dublet, M = 3 triplet, M = 4 quartet etc.
Remark: The spin of the inner electrons does not have to be considered. They cancel each other because
of their opposite algebraic sign in double occupied orbitals.
e)
Part c) and d) demonstrate that dioxygen is a diradical. The Lewis structure does not show any unpaired
electrons. Thus it does not show the correct distribution of electrons.
Solution to problem 1-4
a)
b)
Helium, neon, argon
P4 + 5 O 2
P4O10
1
/8 S8 + O2
SO2
c)
Examples (R = Alkyl group)
KZ 1:
O
C
C
acidic reaction
acidic reaction
KZ 2:
O
O
O

N
O
N
R
basic reaction
basic reaction
Cl
H
O
,
H
,
R
H
Cl
H
+
O
H
O
O
O
2 Li2O
2 CaO
KZ 3:
R

4 Li + O2
2 Ca + O2
H
d)
e)
i)
Oxidation number
Examples
Oxidation number
Examples
-I
H2O2, Na2O2
+½
O2PtF6
-½
KO2
+I
O2F2
1
- /3
NaO3
+II
OF2
0
HOF
2 Fe(OH)3
T
Fe2O3 + 3 H2O
51
Answers Round 1
ii)
Ammonium nitrate shows an acidic reaction. It should remove absorbed hydroxide ions and wash
the precipitate to be neutral. In the presence of chloride ions iron(III) chloride may evaporate and
thus the result is falsified.
iii)
The filter shall be burned without any residue which could be weighed with Fe2O3 to give a wrong
result.
iv)
Fe3O4 is magnetic and could be identified with a magnet.
v)
M(Fe2O3) = 159.69 g/mol
M(Fe3O4) = 231.54 g/mol
1 mol of iron(III) ions form
or

vi)
159.69g/2 = 79.855 g Fe2O3
231.54/3 g = 77.18 g Fe3O4, respectively
If Fe3O4 is formed the mass after annealing is too low. Thus the formation of Fe3O4 leads to a
content of iron which is too low.
Mean result of Fe2O3 after annealing:
(0.2483 g + 0.2493 g + 0.2488 g) : 3 = 0.2488 g .
Amount of iron in 50 mL solution:
n(Fe)50 =
0.2488 g ·2
M(Fe2 O3 )
=
0.2488 g ·2
159,69 g/mol
-3
= 3.116 · 10 mol
 Mass of iron(III) chloride in 250.0 mL:
-3
m(FeCl3) = 5 · 3.116 · 10 mol · M(FeCl3) = 0.01558 mol · 162.21 g · mol
–1
m(FeCl3) of the sample = 2.527 g.
Solution to problem 1-5
a)
MeO2 + 4 HCl
–
Cl2 + 2 I
I2 + 2 Na2S2O3
b)
Starch solution is added to realize the change more easily. Iodine forms a clathrate with starch.
c)
The oxygen of the air can oxidize iodide to iodine. Thus the consumption of thiosulfate would be too high
because more iodine has to be titrated.
In contact with steam chlorine might disproportionate into chloride and hypochlorite. Even if you assume
that the equilibrium of the formation of hypochlorite in the presence of hydrochloric acid lies on the side
of chlorine it cannot be excluded that some hypochlorite forms. As this reaction takes place in the vessel
with the metal oxide chlorine set free by the metal oxide would be lost to oxidize iodide to iodine. This
would lead to a smaller consumption of sodium thiosulfate.
d)
c(thiosulfate) = 0.1 mol /L = 0.1 mmol/mL
24.25 mL of thiosulfate solution ≙ 0.1 mol/mL · 24.25 mL/2  1.21 mmol I2
 290 mg of MeO2 produce 1.21 mmol of I2
1 mmol of MeO2 ≙ 1 mmol of I2
 1.21 mmol of MeO2 ≜ 290 mg
1 mol of MeO2 ≜ 290 g/1.21 = 239.7 g
M(Me) = (239.7 - 2 · 16.00) g/mol = 207.7 g/mol.  Me = Pb
52
MeCl2 + Cl2 + 2 H2O
–
I2 + 2 Cl
–
+
2–
2 I + 4 Na + S4O6
Answers Round 2
Answers Round 2
Solution to problem 2-1
a)
M: Pb,
X: KI,
Y: PbI2,
Z: KPbI3
There are several ways to solve this problem. Only one of these will be described here.
Specification of M: The information about the solubility in acids indicates a passivation in the presence of
sulfate and chloride ions. These facts point to lead which forms a sparely soluble sulfate and a sparely
soluble chloride.
The information about the solubility in sodium hydroxide solution strongly limits the number of metals as
only a few metals can be dissolved in this solution (examples: beryllium, aluminum, tin, iron, lead).
The information of the pyrophoric property limits the selection, too. Metals which can ignite when finely
dispersed are magnesium, titanium, nickel, cobalt, iron and lead as well as rare earth metals of the inner
transition series.
When a sodium hydroxide solution is added, the aqueous solution of the cations forms an insoluble hydroxide which dissolves in an excess of it. Examples are aluminum, lead, zinc, beryllium, chromium, gallium, indium, copper and gold.
With ammonia there is no formation of an ammine complex: iron(III), aluminum, beryllium, titanium,
zinc, tin, lead.
Finding of X and M: The endothermic solvation hints to compounds which are asked for in part f). Compound X must be able to react with the gaseous compound G. Possibilities are the formation of carbonates (from CO2), oxides (from O2), triiodates (from I2). An aspect in favor for the latter is the formation
of a metallic mirror when G is formed (reducing properties of iodine).
X or parts of X and the metal M must form a sparely soluble compound Y which has to have broader
properties: layer structure, properties of a semiconductor, thermochromism. Possibilities are sparely soluble sulfates, chlorides, iodides.
Taking all statements into account only lead is remaining. Y must be lead(II) iodide. X may be sodium or
potassium iodide (see f)).
Part e) hints to a double salt. With the formation of the simplest double salt XPbI2 can the alkali metal be
determined by means of a calculation of the given mass ratio.
b)
c)
2 KI + Pb(NO3)2
PbI2 + 2 KNO3
PbI2 + KI
KPbI3
H = KI3
KI + I2
KI3
I
K
+
–
I
I
–
H is an ionic compound so the cation and the anion are shown separately. In the I 3 anion the central
iodine atom is surrounded by 5 electron pairs resulting in a trigonal bipyramidal arrangement. There are
two bonded and three lone pairs. The lone pairs are equatorial arranged (according to VSEPR it is an
AX2E3 system). All these facts result in a linear structure of H.
d)
The TG plot shows a difference in mass of 5 – 6 %. The molar mass of the water free compound is
M(KPbI3) = 627 g/mol. Then the mass loss is 627 g/mol · 5.5/94.5 = 36.5 g/mol  n = 2.
e)
As the oxidation numbers and the elements are the same as in Z the wanted compound can only be
K2PbI4 with a mass ratio of lead of 207.2/793 · 100 % = 26.13 %
f)
NH4Cl, KI, CaCl2 · 6 H2O, Na2SO4 · 10 H2O, NaI, KNO3, NaNO2.
53
Answers Round 2
The lattice energy is needed in order to dissolve a salt, the solvation energy is set free. As soon as the
needed energy (-lattice energy) is higher than the solvation energy the solution provides the difference
and cools down. Condition for cooling-down:
Lattice energy < Solvation energy or |Lattice energy| > |Solvation energy|
g)
+I,+II,–I
+I,–II
KPbI3 · n H2O
h)
+I,–II +I,–I 0
T
0
n H2O + KI+ Pb + I2
Pb(NO3)2 + 2 NH3 + 2 H2O
Pb(OH)2 + 2 NH4NO3
or as ionic equation:
2+
–
Pb (aq) + 2 OH
Pb(OH)2(s)
Remark: Lead hydroxide is formed. There is no formation of an ammine complex in an aqueous solution.
Solution to problem 2-2
a)
In the reaction to A two isomers are formed. Acetylacetone is deprotonated by triethylamine to yield a
carbanion which is stabilized by mesomerism to an enolate. In doing so the two isomers are formed,
which are trapped by TMSCl.
free rotatability
Afterward the two isomers are deprotonated by LDA and again a carbanion or rather an enolate is
formed which is trapped again by TMSCl:
Both isomers of B react with NBS to yield C. The reaction takes place at the outer double bond.
54
Answers Round 2
b)
E/Z-Isomers are formed:
c)
Assignment and ratio of isomers:
δ = 5.24 ppm: 1 proton at position 2 (Isomer 1)
δ = 4.95 ppm: 1 proton at position 2' (Isomer 2)
Remark: The assignment to isomer 1 or isomer 2 is arbitrary and could be done the other way round but
it should be the same as in the following signal assignments. The signal at 5.24 refers to the Z-isomer, that
at 4.95 to the E-isomer but the students are not expected to know that.
From the ratio of isomer 1 to isomer 2 you find 2.11:1.
The signals of the hydrogen atoms of the methyl groups at position 3 and 3' are almost identical as well as
those of the methyl groups at position 1 und 1':
δ = 1.93 ppm: 6 protons at 3 and 3'
δ = 1.79 ppm: 6 protons at 1 and 1' .
The remaining signals belong to the hydrogen atoms of the trimethylsilyl groups.
δ = 0.00 ppm: 9 protons of the Si(CH3)'3 group
δ = -0.04 ppm: 9 protons of the Si(CH3)'3 group
Calculation of the intensities:
The protons at positions 3 and 3‘ as well as those at 1 and 1‘ are so similar that their signals are almost
identical and thus the signals appear as a singlet.
 The intensity for both singnals at δ = 1.93 ppm and  = 1.79 ppm is 3 · 2.11 + 3 · 1 = 9.33
For isomer 1 the intensity of 9 · 2.11 = 18.99 is expected
for isomer 2 the intensity of 9 · 1 = 9 is expected.
 δ = 0.00 ppm: 9 · 1 = 9 (Isomer 2)
δ = 0.04 ppm: 9 · 2.11 = 18.99 (Isomer 1)
79
81
d)
Compound C contains a bromine atom. Natural bromine consists of two isotopes, Br und Br (=2),
with nearly the same abundance. Thus in the mass spectrum two nearly identical peaks with =2 are
found at 177.96 and 179.96.
e)
Et3N is used as a base. It deprotonates acetylacetone at the CH2 group between the two carbonyl groups,
so an enolate can be formed. This enolate reacts with TMS-Cl to yield TMS-enolether.
f)
The base strength of Et3N does not suffice for a second deprotonation. Thus the stronger Base LDA is
used.
g)
The reaction of acetylacetone with NBS in the presence of a base is not regiospecific and would not lead
to the desired product because the deprotonation of acetylacetone would take place at the CH 2 group
between the two carbonyl groups.
The synthesis via A and B guarantees that only the desired regiospecific compound forms.
55
Answers Round 2
h)
The H/D exchange may happen up to three times so that a completely deuterized CD 3 group may be obtained.
Solution to problem 2-3
a)
Cyclobutadiene
2 π electrons (charge: +2):
4 π electrons (charge: 0):
1P
1P
Remark: The degeneracy of 2 and 3 leads to a
diradical electron structure.
6 π electrons (charge: –2):
Tropylium ions
6 π electrons (charge: +1):
8 π electrons (charge: –1):
Remark: The degeneracy of 2 and 3 leads to a
diradical electron structure.
56
Answers Round 2
The calculation of the total energy is done as follows:


The energy of the π system is compared with the aliphatic version. Therefore the respective energies
of the occupied orbitals have to be calculated with the given formulae and added.
The energy levels of the Frost circle result from the following considerations:
 The height of the respective energy level with respect to the AO level (E = ).
 The difference from the origin to the energy level is 2β.
 The angle  is 360°/n; the angle between the energy level and the height amounts to m · , where
m = number of the energy level – 1 and n = number of carbon atoms in the n-cyclic system.
 To find the height the cosine function is used:
cos(m · ) = H/2β
 H = cos(m · ) · 2β
 
E=+H
The calculation for cyclobutadiene gives = 90°.
Number of electrons
Bond energy
Aliphatic
Difference
2
E = 2 · (+2β)
= 2.0+4.0β
E = 2.0+3.2β
ΔE = 0.8β
4
E = 2 · (+2β) + 2 ·  = 4.0+4.0β
E = 4.0+4.5β
ΔE = –0.5β
6
E = 2 · (+2β) + 4 ∙  = 6.0+4.0β
E = 6.0+3.2β
ΔE = 0.8β
 The cyclobutadienyl dication and the cyclobutadienyl-dianion are aromatic because of E > 0.
The calculation for tropylium gives  = 51.429°.
Number of electrons
Bond energy
Aliphatic
6
E = 2 · (+2β) + 4 · (+cos () ∙ 2β) = 6.0+9.0β
E = 6.0+8.1β
8
E = 2 · (+2β) + 4 · (+cos() · 2β) + 2 · (+cos(2) · 2β)
= 8.0+8.1β
E = 8.0+8.1β
 The tropylium cation is aromatic because of E > 0.
b)
The compound has to be
1. cyclic,
3. totally conjugated
Difference
ΔE = 0.9β
ΔE = 0
2. planar,
4. fulfill the Hückel rule: it must have 4n + 2 π- electrons.
c)
The following compounds (ions) fulfill all conditions and are aromatic:
(i) Pyrrole (6 π electrons), (iii) azulene (10 π electrons), (v) pyridinium cation (6 π electrons), (vi) caffeine
(10 π electrons and 6 π electrons in the five membered ring).
Non-aromatic ore antiaromatic are:
(ii) Allyl anion (4 π electrons; violates rules 4 and 1), (iv) 1H-pyrrolium cation (N protonated pyrrole) (violates rule 3)
d)
HOMO: 2 and 3; LUMO: 4 and 5
Remark: The orbitals are degenerate (having the same energy).
57
Answers Round 2
e)
The decision is based on the number of nodal planes in the bond axis of the respective orbital. Orbitals of
same symmetry show interaction.
HOMO (Cp ), (2,3 )
–
LUMO (Cp ), (4,5)
–
–
–
x
x
–
x
x
–
x
–
x
–
–
–
–
–
–
3dx2-y2
3dz2
3dxy
3dxz
3dyz
4s
4px
4py
4pz
–
x = interaction expected
- = no interaction expected
Orbital interaction between
–
Fe and Cp is possible in each
column.
5
5
f)
[Ru(η -Cp)2] is more stable as it fulfills the 18 electron rule. [Rh(η -Cp)2] has 19 electrons and should act
as a reducing reagent.
g)
Possible structures:
18 Valence electrons each
18 Valence electrons each
18 Valence electrons each
Remark: The numbers of valence electrons at the metal centers are found as the sum of all electrons assigned to the metal. Each bond of the terminal CO groups counts for 2 electrons, the bridging CO groups
provide 1 electron for each metal center and the cyclopentadienyl ring 6 electrons. The 7 d-electrons of
the metal centers are added  Σ = 18.
Solution to problem 2-4
a)
2+
+
[Cu(H2O)5(OH)] + H3O
c2
c3
M(Cu(NO3)2) = 187.56 g/mol
 c(Cu(NO3)2) = 8.96 · 10 mol/L.
Ka =
–4.40
–3
–3
mol/L
c1 + c2 = 8.96 · 10 mol/L
–3
c1 = 8.96 · 10 mol/L – 10
–4.40
–3
mol/L = 8.92 · 10 mol/L

–7
Ka = 1.78· 10
58
(The tetraaqua complex is also accepted)
c2 /(1 mol/L) · c3 /(1 mol/L)
c1 /(1 mol/L)
c2 = c3 =10
b)
+
[Cu(H2O)6] + H2O
c1
pKa = 6.75
Ksp. 25°C = c(Cu )·c (OH )/(1 mol/L)3 
2+
2
–
Ksp .25°C · (1 mol/L)3
c(OH ) = √
–
c(Cu2+ )
Answers Round 2
c(OH ) = √
–
1.6 ·10−19
1.03 ·10−2
–
mol/L
 pOH = 8.40
–9
c(OH ) = 3.94 · 10 mol/L
pH = 5.6
2+
At this pH the fraction of protolysed Cu ions can be neglected.
c)
2 Cu
+
2+
Cu + Cu
o
E1 = E 1 +
RT
F
+
o
· ln c(Cu )
E2 = E 2+

In equilibrium: E1 = E2.
ln K = (0.52 – 0.16) ·
RT
F
2+
+
· ln (c(Cu ) / c(Cu ))
o
F
o
ln K = (E 1 – E 2 ) ·
96485
= 14.16
8.314·295
RT
T = 295 K
K = 1.41 · 10
6
d)
According to the electron configuration copper(I) should be most stable because it has a closed electron
10
9
shell (d configuration). In contrast copper(II) has a d system. The stability of copper(II) in aqueous
solutions is caused by the high hydration enthalpy which is set free when the small, highly charged
copper(II) ions are dissolved. This effect overbalances the unfavorable electron configuration.
e)
K=
c(Cu2+ )/(1 mol/L)2
c2 (Cu+ )/(1mol/L)
+
2+
c(Cu ) = x mol/L
K = y/x
c(Cu ) = y mol/L
x + 2 y = 0.01 
2
K = (0.01 – x) / 2 x
2
y = ½ (0.01 – x)
2
x + (x / 2 · K) – (0.01 / 2 · K) = 0
6
f)
K = 1.72 · 10
–5
+
–5
–3
2+
–3
x = 5.38 · 10  c(Cu ) = 5.38 · 10 mol/L,
y = 4.97 · 10  c(Cu ) = 4.97 · 10 mol/L
Copper(I) oxide can only dissolve when disproportionation takes place. This will not be if E2 > E1 (see c)).
+
–
+
–15
–
+
+
c(Cu ) = KL· (mol/L)²/ c(OH )
 c(Cu ) = 10 · (mol/L)²/c(OH )  c(Cu ) = 0.1 · c(H3O )
2+
c(Cu ) = 0.01 mol/L
RT
RT
+
· ln [0.1 · c(H3O+)/( mol/L)]
E1 = 0.46 + · ln [c(H3O )/(mol/L)]
F
F
RT
RT
+
+
E2 = 0.16 + · ln[0.01 / (0.1 ·c(H3O )/(mol/L))]
E2 = 0.10– · ln [c(H3O )/(mol/L)]
F
F
−0.36 ·F
+
+
E2 > E 1 
ln [c(H3O )/(mol/L)] <
ln [c(H3O )/(mol/L)] <–7.13
2 ·R ·T
E1 = 0.52 +
+

–4
c(H3O ) < 8.01 · 10 mol /L
pH = 3.1
The higher the temperature the lower the pH has to be.
+
ln [c(H3O )/(mol/L)] <
Temp.
c(H3 O)+
ln
1 mol/L
in °C
0
10
20
30
40
50
60
70
80
90
100
-7.65
-7.38
-7.13
-6.89
-6.67
-6.47
-6.27
-6.09
-5.92
-5.75
-5.60
c(H3 O)+
1 mol/L
-4
4.76·10
-4
6.24·10
-4
8.01·10
-3
1.02·10
-3
1.27·10
-3
1.55·10
-3
1.89·10
-3
2.27·10
-3
2.69·10
-3
3.18·10
-3
3.70·10
−0.36 ·F
2 ·R ·T
you get the following table:
pH
3.32
3.20
3.10
2.99
2.90
2.81
2.72
2.64
2.57
2.50
2.43
3.3
3.1
pH value
With
2.9
2.7
2.5
2.3
Temperature in °C
59
Answers Round 3 Test 1
Answers Round 3 Test 1
Solution to problem 3-01
A
a)
2 Na2S2O3 +1 I2
b)
2
c)
1
2 NaI + 1 Na2S4O6
2+
–
–
Ba + 2 MnO4 +1 CN
–
+
–
ClO3 + 6 H3O + 6 Br
+ 2 OH
-
–
2 BaMnO4 + 1 CNO + 1 H2O
–
3 Br2 + 1 Cl + 9 H2O
B
2+
Fe in acidic solution
3+
Al in acidic solution
2+
Cu in acidic solution
2+
Cu in ammoniac solution
Cl in basic solution
+
Na
light green
colorless
blue
deep blue
colorless
colorless
Iron sulfide (FeS)
Copper sulfate (CuSO4)
Silver iodide (AgI)
Potassium sulfate (K2SO4)
Potassium permanganate (KMnO4)
Potassium chromate (K2CrOI4)
C
Silver bromide, lead sulfate, iron(II) sulfide.
D
Ba(NO3)2, K2(COO)2, Al2O3, K2MnO4, KAl(SO4)2, Na2CO3·10H2O
E
Soluble are potassium, aluminum, zinc.
F
Soluble are potassium, lead, copper, zinc.
black
white
yellow
white
violet/black
yellow
Solution problem 3-02
a)
Cu + 2 A g  Cu + 2 Ag
+
-8
(Only if c(Ag ) ≤ 1.7∙10 mol/L the reaction runs in the reverse direction.)
+
2+
b)
U/V
0.45
0.40
0.35
0.30
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
log x
c)
(Remark: The electrode at which a positive charge carrier goes from the metal into the solution or a negative charge carrier comes out of the solution on to the metal (oxidation) is called anode, in the opposite
case it is the cathode.)
+
Cathode: 2 Ag + 2 e
 2 Ag
2+
Anode: Cu
 Cu + 2 e
+
2+
Cell reaction: 2 Ag + Cu  2 Ag + Cu
U = E
E = E(cathode) – E(anode)
E = 0.80 V +
8.314 JK−1 mol−1 ·298 K
96485 Cmol−1
· ln 0.02 – 0.34 V
(or taken from the plot at lg x = -1.70)
60
E = 0.36 V
Answers Round 3 Test 1
d)
At equilibrium:
0.80 V +

R ·T
F
E(cathode) = E(anode)
+
0
· ln(c(Ag )/c ) = 0.34 V +
ln K = ln
c(Cu2+ ) · c0
c(Ag+ )2
=
R ·T
2+
0
· ln(c(Cu )/c )
2·F
(0.80−0.34)𝑉 ·2 · 𝐹
0
with c = 1 mol/L
R ·T
15
e)
ln K = 35.8
K = 3.6 · 10
Determination of the concentration of the silver ions:
E = E(cathode) – E(anode)

0.420 V = 0.34 V – [0.80 V +
R ·T
F
+
0
· ln(c(Ag )/c )]
 c(Ag ) = 1.31·10 mol/L
Determination of the concentration of the iodide ions:
–
m(KI) = 3.00 g
M(KI) = 166 g/mol  n0(I ) = 0.0181 mol
after addition of the silver nitrate solution:
+
+
n(I ) = n0(I ) - n0(Ag )
n0(Ag ) = 0.05 L · 0.200 mol/L = 0.010 mol
-3
n(I ) = 0.0181 mol - 0.010 mol
n(I ) = 8.1 · 10 mol
-3
-2
c(I ) = 8.1 · 10 mol/0.1 L
c(I ) = 8.1 · 10 mol/L
+
0
0
-15
-2
 Ksp = c(Ag )/c · c(I )/c
Ksp = 1.31·10 · 8.1 · 10
-16
Ksp = 1.1 · 10
+
-15
Solution to problem 3-03
a)
b)
c)
1 example of the known compounds
or of the also possible
not possible following the statements:
ClF: 2, BrF: 2, IF: 1, BrCl: 3, ICl: 2, IBr: 2
XY3: T-shaped
d)
e)
ClF3, BrF3, IF3, ClF5, BrF5, IF5, IF7, (ICl3)2
ClF7, BrCl3,5,7, ICl5,7, IBr3,5,7
FCl3,5,7, FBr3,5,7, FI3,5,7, ClBr3,5,7, ClI3,5,7, BrI3,5,7
XY5: square pyramidal
XY7: pentagonal bipyramidal
= X, = Y, ---- = free electron pair
XY + H2O
HY + HXO
3 XY
XY3 + X2
or
5 XY
XY5 + 2 X2
or
7 XY
XY7 + 3 X2
Solution to problem 3-04
a)
g
ZnSO4(s)  ZnSO4(aq):
Enthalpy of solution Hsol (ZnSO4) = -0.900 kJ/K · (22.52 – 22.55) K / 1.565
= -0.873 kJ / 1.565 g
-1
With M(ZnSO4) = 161.48 g/mol:
Hsol (ZnSO4) = 161.48 g mol · (-0.873 kJ / 1.565 g)
Hsol (ZnSO4) = -90.08 kJ/mol
ZnSO4 · 7 H2O(s)  ZnSO4(aq):

b)
Hsol(ZnSO4 · 7 H2O) = -0.900 kJ/K · (21.84 – 22.15) K / 13.16 g
= 0.279 kJ / 13.16 g
-1
With M(ZnSO4 · 7 H2O) = 287.59 g/mol: Hsol(ZnSO4 · 7 H2O) = 287.59 g mol · (0.279 kJ / 13.16 g)
Hsol(ZnSO4 · 7 H2O) = 6.10 kJ/mol
HR = Hsol (ZnSO4) - Hsol(ZnSO4 · 7 H2O) = -90.08 kJ/mol - 6.10 kJ/mol
HR = -96.18 kJ/mol
HR of reaction
NH4NO2(aq)
½ H2(g) + ½ N2(g) + O2(g)  HNO2(aq)

NH3(aq) + HNO2(aq)
has to be calculated:
- H4
61
Answers Round 3 Test 1
+
+
+
+




NH4NO2(s) + aq
N2(g) + 2 H2O(l)
NH3(aq)
2 H2(g) + O2(g)
NH4NO2(aq)
NH4NO2(s)
½ N2(g) + 1.5 H2(g)
2 H2O(l)
+ H5
- H1
- ½ H3
+ H2
+ aq
½ H2(g) + ½ N2(g) + O2(g)  HNO2(aq)
- H4 + H5 - H1 – ½ H3 + H2
HR = (38.1 + 25.1 + 307.4 + 161.7/2 – 571.7) kJ/mol = -120.3 kJ/mol
Solution to problem 3-05
a)
Volume of the cube: V = a
3
Volume of the sphere V =
fcc
4  r3
3
= 4,2 r
3
bcc
Face diagonal of the cube: dface = a
Length of the face diagonal = 4 r
2  a=2r
4r=a
Space diagonal of the cube: d space = a
2
3
Length of the space diagonal = 4 r
2
4r=a
3
3
Volume of the cube = (2 r 2 ) = 22.6 r ≙ 100 %
3
Z = 4  total volume of the atoms = 16.8 r
Filled space = (16.8 / 22.6) · 100 % = 74 %
3  a=4r/
3
3
3
Volume of the cube = (4 r/ 3 ) = 12,3 r ≙ 100 %
Z = 2  total volume of the atoms = 8.4 r
Filled space = (8,4 / 12.3) · 100 % = 68 %
74−68
3
· 100 %  8 %.
b)
The difference in density for the same kind of atoms is about
c)
2 LiAlSi2O6 + 4 CaCO3
d)
Due to its small size and its charge the lithium cation solvated stronger than the other alkali metals. Thus
it dissolves in polar, non-aqueous solvents, too.
e)
When potassium aluminum sulfate is dissolved hydrated aluminum cations form. These tend to hydrolysis
and aluminum hydroxide is formed which can be dissolved again by a solution of sodium hydroxide:
3+
[Al(H2O)6]
f)
Li2O + 4 CaSiO3 + Al2O3 + 4 CO2
2+
[Al(H2O)5(OH)] + H
[Al(H2O)3(OH)3] + OH
74
–
+
+
[Al(H2O)4(OH)2] + 2 H
+
[Al(H2O)3(OH)3] + 3 H
+
–
[Al(OH)4] + 3 H2O
Otherwise there is the danger that aluminum hydroxide precipitates and falsifies the result.
Remark: At pH = 12.6 the concentration
of dissolved aluminum has the highest
value.
2–
3+
–
+
g)
H2Y + Al
h)
During the complexation with EDTA protons are set free. An excess of protons in the progression of the
titration would shift the equilibrium in the direction of the non-complex metal (or the metal-EDTA complex is less stable).
i)
EDTA – Al > EDTA – Zn > Ind – Zn > Ind – Al
62
3+
AlY + 2 H
Dependency of the aluminum(III) concentration
of a saturated aqueous solution of pH 25 °C
2+
2+
3+
Answers Round 3 Test 1
j)
Consumption of Na2EDTA = (50.00 – 15.25) mL = 34.75 mL
 n(Al) = 3.475 mmol
n(Al2O3) = ½ n(Al)  m(Al2O3) = ½ n(Al) · M(Al2O3)
m(Al2O3) = ½ · 3.475 mmol · (101.96 mg/mmol) = 177.16 mg
m(Li2O) = (198.0 – 177.16) mg = 20.84 mg
n(Li2O) = m(Li2O) / M(Li2O)
 n(Li2O) = 20.84 mg / (29.88 mg/mmol) = 0.70 mmol
½ · 3.475 mmol Al2O3 : 0.70 mmol Li2O  2.48 : 1  5 : 2

x = 2 und y = 5
Empirical formula: 2 Li2O · 5 Al2O3
Solution to problem 3-06
a)
b)
log Kp = -4374/473+ 1.75·log 473 + 3.78
Kp =
p(PCl3 )/p0 ·p((Cl2 )/p0
p(PCl5 )/p0
log Kp· = -0.786
0
with p = 1.000 bar
Let x = p(PCl3)/1 bar:
0.200 =
x2
1.5 −2 ·x
Kp = 0.164
p(PCl3) = p(Cl2) und p(PCl5) + 2· p(PCl3) = 1.50 bar

2
x + 0.400 ·x 0150 – 0.300 = 0
x1 = 0.383 (x2 = -0.783)
p(PCl3) = p(Cl2) = 0.383 bar = 38.3 kPa
p(PCl5) = 0.734 bar = 73.4 kPa
c)
p(PCl3 )
p(PCl5 )
=
n(PCl3 )
n(PCl5 )
=
383
(1)
734
In equilibrium: n(PCl5) = n0(PCl5) – n(PCl3).
Assume that n0(PCl5) = 1 mol
 n(PCl3) = 1 mol - n(PCl5) and inserted in (1):
n(PCl3 )
n(PCl5 )
d)
=
1 mol− n(PCl5 )
n(PCl5 )
=
383

734
(1 +
) · n(PCl5) = 1 mol
734
n(PCl5) = 0.657
i.e. (1 - 0.657) mol of PCl5= 0.343 mol of PCl5 ≙ 34.3 % of the original PCl5 have reacted.
+
–
PCl5 as a solid consists of PCl4 and PCl6 ions:
+
Cl
Cl P
Cl
Cl
P
Cl
Cl
–
Cl
Cl
e)
383
Cl
Cl
There must be a rapid exchange between the axial and the equatorial position which is faster than the
NMR experiment. Thus only one signal is detected.
Remark: The exchange of the positions takes place in the so called Berry pseudoration. Both, the axial
and the equatorial constituents move at the same rate of increasing the angle between the other axial or
equatorial constituent. This forms a square based pyramid. The frequency of the rotation for PF5 is about
–1
100000 s (298 K) and cannot be detected by the slower NMR experiment.
X = F, Cl
63
Answers Round 3 Test 1
Solution to problem 3-07
O
Systematic name: Propan-2-one
Other names: Acetone, Propanone, 2-Propanone, dimethyl ketone, β-Ketopropane
OH
Systematic name: prop-1-en-2-ol
Other names: 1-Propen-2-ol, propen-2-ol,2-hydroxypropene
OH
1-Propen-1-ol
OH
2-Propene-1-ol, allyl alcohol, 2-propenol
O
CH3
Systematic name: Epoxypropane
Other names: 1,2-Epoxypropan, 2-Methyloxiran, propylene oxide, propene oxide,
1,2-propylene oxide, methyl oxirane, methyl ethylene oxide, methylethylene oxide
OH
Systematic name: Cyclopropanol
Other names: Cyclopropyl alcohol, hydroxy cyclopropane
Systematic name: Oxetane
Other names: oxacyclobutane, trimethylene oxide, 1,3-propylene oxide,
1,3-epoxypro-pane
Systematic name: Methyl vinyl ether
Other names: methoxyethene, ethyl methyl ether
O
O
O
Systematic name: Propanal
Other names: Propionaldehyde
Solution to problem 3-08
a)
A:
*B:
*C:
E:
F:
G:
*i: AlCl3
*ii: CH3Cl
iii: CH3COCl
iv: HNO3/
H2SO4
v: H2SO4/
HNO3
D:
vi: HCl
vii: Br2
* there are more possible solusions
Isomerization: Keto-enol tautomerism.
b)
64
During the formation of G the temperature must be kept below 5 °C. When heated to boiling 3nitrophenol is formed:
Answers Round 3 Test 1
Remark: A diazonium coupling does not occur because diazonium salts can only attack
strongly activated aromatic compounds. The aromatic compound at hand however is deactivated because of the strong –M effect of the NO2 group.
c)
The amino group shows a +M effect and is ortho/para directing.
For the directing effect of the substituent see organic text books
The meta-substituted product does not match to the +M effect of the amino group. Under these strongly
acidic reaction conditions the amino group is protonated and therefore without a free electron pair. Thus
the +M effect cease to exist and the –I effect works only which leads to a preference of the meta substitution.
Solution to problem 3-09:
a)
Electrophilic addition (AE) of Br2 at the double bond of the butenone:
Bromonium cation
b)
Nucleophilic addition (AN) of HCN at the keto group and formation of a cyanohydrin:
c)
There is a resonance structure in which the β-C atom is positively charged. There the Grignard reagent
may perform a nucleophilic attack:
d)
The C atom directly bonded to oxygen is harder because the oxygen attracts the electrons of the C=O
bond and thus the carbonyl carbon is less polarizable. The β-carbon atom, however, is less concerned by
the electron attraction of the oxygen atom and thus more polarizable, weaker.
The hard Cer reagent prefers to attack the harder carbonyl group, the copper reagent the weaker βcarbon.
65
Answers Round 3 Test 2
Answers of Round 3 Test 2
Solution to problem 3-11
a) E
b) (Chromium has the oxidation number +6) D, E (Peroxodisulfuric acid)
c) B
d) B, E
e) B, D, E
h) A, C, D
f) B
g) B, C (cis-trans), E
Solution to problem 3-12
c(HA)
c(A− )

c(HA)
= 10pKs−pH
c(A− )
a)
pKS = pH + lg
b)
HCOOH + NaOH  HCOONa + H2O
c(HCOOH) = 0.1 mol/L – c(NaOH)
pKS = pH + lg
= 10
6.5 – 5.8
–
and
0.1 mol/L− c(HCOO− )
c(HCOO− )
=5:1
c(HCOO ) = c(NaOH)

c(HCOO ) = (0.1 mol/L) / (10pKs−pH + 1)

pKS(NH4 ) = 9.25
–
–
c(HCOO ) = 0.074 mol/L
c)
0.1 mol/L
0.1 mol/L
+
It is pKS(NH4 ) = pH + lg
-10.25
It has to be
10
+
- lg KS = - lg c(H ) + lg
+
+
-8.25
< c(H ) < 10
0.1 mol−n(NaOH)
0.1 mol+n(NaOH)
Addition of NaOH solution:
-10.25
10
0.1 mol−n(NaOH)
0.1 mol+n(NaOH)
-9.25 0.1 mol+n(NaOH)
10
·
0.1 mol−n(NaOH)

<
+
c(H ) = KS ·
0.1 · (0.1 mol + n(NaOH) < 0.1 mol - n(NaOH)
n(NaOH) < 0.0818 mol
V(NaOH) = n/c < 81.8 mL
The following calculation can be left out and the result immediately given because it is a symmetrical
problem.
-8.25
Addition of HCl solution:
10
-9.25
< 10
·
0.1 mol−n(HCl)
0.1 mol+n(HCl)
n(HCl) < 0.0818 mol
d)
Equation for the freezing-point depression:
T = K ·
n
m(solvent)
V(HCl) = n/c < 81.8 mL
n = amount of all dissolved particles
Determination of the cryoscopic constant of water:
K = T·
m(solvent)
m(sucrose)/M(sucrose)
K = 0.052 K ·
0.1 kg
0.957 g/(342.3 g/mol )
-1
K = 1.86 K·kg·mol
Determination of the carboxylic acid X:
+
–
HA
H +A
+
–
n = n(HA) + n(H ) + n(A )
0.147 K = 1.86 K·kg·mol
It is a weak acid
 M(X) = 73.3 g/mol
X = R–COOH
 R = CH3–CH2
e)
T = K ·
-1
2.895 g ·(1+ )
·
M(X) · 0.5 kg
 (1 + )  1
M(X) = 73.3 g/mol = (28.3 +45) g/mol
X = propionic acid with M = 74.08 g/mol
n0 ·(1+)
m(solvent )

n0 = 2.895 g /(74.08 g/mol)
=
66
n = n0·(1 - ) + n0· + n0· = n0·(1 + )
0.147 K · 0.5 kg · 74.08 g·mol−1
1.86 K·kg·mol−1 ·2.895 g
=
T ·m(solvent)
-1
K · n0
K = 1.86 K·kg·mol
-1
-1
 = 0.0112
T = 0.147 K
m(solvent) = 0.5 kg
Answers Round 3 Test 2
KS =
KS =
c0 ·  · c0 · 
c0 (1− )
0
/c
KS = c0 ·
2.895 g ·0.01122 ·1 L ·mol−1
74.08 g ·mol−1 ·0.5 L · (1−0.0112)
2
1− 
0
0
/c
c = 1 mol/L
500 g of water ≙ 0.5 L of water
-6
KS = 9.92 · 10
Solution to problem 3-13
a)
Solution of 

NH4Cl
BaCl2
Na2S
Pb(NO3)2
Na2SO4
AgNO3
n. r.
weak gas
formation
white prec.
n.r.
white prec.
white prec.
white prec.
white prec.
white prec.
black prec.
n.r.
black prec.
white prec.
n. r.
NH4Cl
BaCl2
Na2S
Pb(NO3)2
Na2SO4
n. r.
AgNO3
b)
NH4Cl (aq) + Na2S (aq)
NH3 (g) + NaCl (aq) + NaHS (aq)
2 NH4Cl (aq) + Pb(NO3)2 (aq)
2 NH4NO3 (aq) + PbCl2 (s)
NH4Cl (aq) + AgNO3 (aq)
NH4NO3 (aq) + AgCl (s)
BaCl2 (aq) + 2 AgNO3 (aq)
Ba(NO3)2 (aq) + 2 AgCl (s)
BaCl2 (aq) + 2 Na2S (aq)
Ba(OH)2 (s) + 2 NaCl (aq) + 2 NaHS (aq)
BaCl2 (aq) + Pb(NO3)2 (aq)
Ba(NO3)2 (aq) + PbCl2 (s)
BaCl2 (aq) + Na2SO4 (aq)
BaSO4 (s) + 2 NaCl (aq)
Na2S (aq) + Pb(NO3)2 (aq)
PbS (s) + 2 NaNO3 (aq)
Na2S (aq) + 2 AgNO3 (aq)
Ag2S (s) + 2 NaNO3 (aq)
Pb(NO3)2 (aq) + Na2SO4 (aq)
PbSO4 (s) + 2 NaNO3 (aq)
Die Lösungen lassen sich eindeutig identifizieren, da sich das Reaktionsverhalten bei allen unterscheidet,
wie aus der folgenden Zusammenstellung sichtbar wird.
Solution of
NH4Cl
BaCl2
Results of the
experiments
1 x gas
2 x white prec.
4 x white prec..
Na2S
1 x gas
1 x white prec..
2 x black prec.
Pb(NO3)2
Na2SO4
AgNO3
3 x white prec.
2 x white prec.
2 x white prec..
1 x black prec.
1 x black prec.
Solution to problem 3-14
a)
G = H – T·S
Cp = -7.8
H(T) = H(298 K) + Cp·(T – 298 K)
-1
-1
H(873 K) = -198 kJ/mol – 7.8 JK mol ·575 K
-1
H(873) = -202.49kJ mol

G(873) = -31.92 kJ/mol
S(T) = S(298 K) + Cp·ln(T/298K)
-1
-1
S(873) = -187 JK mol – 7.8·ln(873 K/298 K)
-1
-1
S(873) = -195.38 JK mol
–(G/RT)
Kp = e
K(873) = 81.27
b)
By doing so the formation of nitrogen oxides is avoided.
c)
Assumption:
100 mol mixture at the beginning
x: reacted amount of O2 in mol
o
5
Kp = 65.00
p = 1.000·10 Pa
67
Answers Round 3 Test 2
SO2
O2
SO3
N2

10
11
0
79
100
Amount at equilibrium
10 – 2x
11 – x
2x
79
100 – x
Molar fraction at equilibrium
10− 2𝑥
100−𝑥
11−𝑥
100−𝑥
2𝑥
100−𝑥
79
100−𝑥
1
10−2𝑥
0
·p
100−𝑥
11−𝑥
0
·p
100−𝑥
2𝑥
0
·p
100−𝑥
Amount at the begin
Partial pressure at equilibrium

65.00 =
79
100−𝑥
· p0
4 𝑥 2 ·(100−𝑥)
(10−2 𝑥)2 ·(11−𝑥)
 256 x – 5060 x + 35100 x – 71500 = 0
x - 19.77 x + 137.11 x – 279.30 = 0
x  3.5 (solved by iteration)
Amount at equilibrium: (100 – x) mol = 96.5 mol
Ration of volume
SO2: (10 – 2·3.5)/96.5·100% = 3.1 %
SO3: 7.3 %
O2: 7.8 %
N2: 81.9 %
Conversion of SO2: 2·3.5/10·100% = 70.0 %
3
2
3
2
Solution to problem 3-15
+
+
+
a)
H represents H3O bzw. H (aq).
–
2+
H2S + 4 Br2 + 4 H2O  8 Br + SO4 + 10 H
+
–
H + OH
 H2O
+
–
-3
-4
n(H ) = n (OH ) = 19.95·10 L · 0.100 mol/L = 19.95·10 mol
-4
+
-5
19.95·10 mol H ≙ 19.95·10 mol H2S
-5
-5
-3
19.95·10 mol H2S in 10.0 mL volcano water ≙19.95·10 mol/0.0100 L = 19.95·10 mol/L
-3
Mass contend = 19.95·10 mol/L · 34.086 g/mol = 0.680 g/L
b)
IO3 + 5 I + 6 H
–
[ggf. I2 + I
I2 + H2S
2I2 + 2 S2O3
–
–
+




3 I2 + 3 H2O
–
I3 ]
–
+
2I +S+2H
–
22 I + S4O6
Amount of prepared iodine (I2):
n(KIO3) =
6.5 g
-4
= 3.04 · 10 mol
100 ·214.0 g/mol
-4
n(I2) = 3 · n(KIO3) = 9.12 · 10 mol
Amount of H2S:
0.0100 L ·0.800 g/L
-4
= 2.35 · 10 mol
34.086 g/mol
-4
-4
2.35 · 10 mol of H2S are oxidized by 2.35 · 10 mol of I2
-4
-4
-4
(9.12 ·10 – 2.35 · 10 ) mol = 6.77· 10 mol of I2 are not consumed
-4
-4
2-3
26.77· 10 mol I2 consume 2 · 6.77· 10 mol S2O3 = 1.354 · 10 mol S2O3 in the titration
-3
-2
V = n/c  V = 1.354 · 10 mol / (0.100 mol/L) = 1.354 · 10 L  13.5 mL
c)
H2S + Ag  Ag2S + H2
Solution to problem 3-16
 2Na + 3 N2
a)
b)
2 NaN3
c)
10 Na + 2 KNO3
 K2O + 5 Na2O + N2(g)
K2O + Na2O + 2 ϑSiO2
 Na2O3Si + Na2O3Si
5
Amount of nitrogen in 50 L at T = 423 K and p = 1.3·10 Pa:
d)
N N N
n(N2) =
68
p·V
R·T
N N N
n(N2) =
(1)
N N N
–
1.3 ·105 Pa · 50 ·10−3 m3
8.314 JK−1 mol−1 · 423 K
(2)
(3)
n(N2) = 1.85 mol
po
Answers Round 3 Test 2
n(N2) = 1.5 · n(NaN3) + 0.1 · n(NaN3) (1) und (2)
1.85 mol = 1.6 · n(NaN3)  m(NaN3) = 1.85 mol/1.6 · 65.0 g/mol
m(NaN3) = 75.2 g
e)
There are more resonance structures but they have a higher separation of charge.
f)
V for ϑ > 100 °C and ϑ < 100 °C have to be considered:
3
Reaction, volume in cm (ϑ > 100 °C/ ϑ < 100 °C)
Hydrogen
Ammonia 1
Ammonia 2
Carbon
monoxide
Ethene
Methane
2 H 2 + O2
20 10
4 NH3 + 3 O2
20
15
4NH3 + 7 O2
20
35
2 CO + O2
20
10
C2H4 + 3 O2
20
60
CH4 + 2 O2
20 40

2 H2O
20/ 0

2 N2 + 6 H2O
10/10 30/ 0

4 NO2 + 6 H2O
20/20
30/ 0

2 CO2
20/20
 2 CO2 + 2 H2O
40/40 40/ 0

CO2 + 2 H2O
20/20 40/ 0
V for ϑ > 100 °C
V for ϑ < 100 °C
V = -10 cm
V =  -30 cm
V = 5 cm
3
3
3
V =  -25 cm
V = -5 cm
V =  -35 cm
3
3
V = -10 cm
3
3
V = -10 cm
3
V = 0 cm
3
V =  -40 cm
V = 0 cm
3
V =  -40 cm
3
3
At ϑ > 100 °C it could be hydrogen, at both conditions carbon monoxide.
Solution to problem 3-17
a)
b)
c)
AsH3  As + 3/2 H2
v = k · c(AsH3)
pend = 1.5 · 86.1 kPa = 129.15 kPa.
-kt
For a reaction of first order it is
c = c0 · e .
-kt
Using c = n/V and p · V = n · R · T you get with constant V and T:
pt(AsH3) = p0(AsH3) · e
 k = ln[pt(AsH3) / p0(AsH3)] / (-t)
k = ln[p120(AsH3) / 86.1 kPa] / (-120 min)
(1)
Determination of p120(AsH3):
p120(AsH3) = p0(AsH3) - x
For each amount of AsH3 which decomposes a 1.5 fold amount of H2 forms:
p0  p120 = p0(AsH3) – x + 1.5 · x = p0(AsH3) + 0.5 · x
 86.1 kPa + 0.5 · x = 112.6 kPa
x = 53.0 kPa
 p120(AsH3) = 86. 1 kPa – 53.0 kPa = 33.1 kPa
(1)
k = ln[33.1 kPa / 86.1 kPa] / (-120 min)
Half-life: ½ p0 = p0 · 𝑒 −𝑘· 𝑡1/2
d)
0.01 = 𝑒
−1.3 ·10−4 𝑠 −1 · 𝑡99%


t1/2 = ln 2 / k
k = 0.00797 min
-1
t1/2 = ln 2 / 0.008 min
-4 -1
t99% = - ln 0.01 / (1.3·10 s )
-1
t1/2 = 87.0 min
t99% = 35424 s t99% = 590 min
Solution to problem 3-18
a)
The structural element of epoxides is a three membered ring which has a high ring strain. Thus ringopening reactions are favored which explains the reactivity of epoxides.
69
Answers Round 3 Test 2
b)
O
H3C
CH3I
S
S
CH3
CH3
I
H3C
Base
-HI
S
CH2
CH3
C
H
O
-S(CH3)2
D
B
c)
1,3 Elimination
d)
e)
E:
F:
H2N
O
G:
OH
H:
OH
I:
J:
OH
HS
Br
R
OH
OH
OH
f)
O
LSkat.
O
H
K
O
LSkat.
H
O
O
L/M
L/M
L/M
M/L
Solution to problem 3-19
a)
70
Only the nitrogen atom with a double bond adjacent to a carbon atom is protonated. The free electron
pair of the other N atom is part of the aromatic system and therefore only weakly basic.
Answers Round 3 Test 2
b)
c)
O
SH
OH
OH
OH
H2N
H2N
O
O
S-Aspartic acid
(S)-Asparaginsäure
d)
R-Cysteine
(R)-Cystein
The OH group of the carboxyl function performs a +M effect on the carbonyl atom and reduces its electrophilic power.
e)
Tetrahedral structure
f)
It is a redox reaction (exactly: oxidative coupling):
O
NH2
HO
SH
+
O
HS
NH2
OH
S
HO
NH2
O
OH
S
NH2
+
H2
O
Solution to problem 3-20
a)
A
Acetaldehyde
B
Acetone
C
Acetophenone
D
Methanoic acid
E
Aniline
F
Benzaldehyde
G
Nitrile of benzoic acid
H
3-Nitrobenzoic acid
I
Phenol
b)
Acidic reaction in water
Reacts with itself in an aldol reaction
is a monosubstituted aromatic compound
Aldehyde as a
functional group
D
A
F
Aromatic compound with metadirecting substituent(s)
H
C
G
–
I
B
E
71
Answers Round 4 (theoretical)
Answers Round 4 (theoretical)
Solution to problem 4-01
a)
b)
- the substance is not decompose producing gaseous products
- the substance is to produce any other volatile components besides water
- the substance is not react with components of the air
- the water has to be released totally
- the content of water is not change under standard conditions
The following reactions may occur because sulfuric acid is strongly acidic and may react oxidizing:
1
H2S + H2SO4
SO2 + /8 S8 + 2 H2O
2HI + H2SO4
I2 + SO2 + 2 H2O
2 NH3 + H2SO4
c)
(potentially 2 H2S + SO2
(NH4)2SO4 (resp. NH3 + H2SO4
3
/8 S8 + 2 H2O)
NH4HSO4)
Calcium chloride forms ammine complexes (ammoniates) of different composition [Ca(NH 3)n]Cl2 (n = 1, 2,
4, 8).
d)
–
2–
+
Isoelectronic species: CO, CN , NO , NN
C C
e)
[Cu(NH3)2]Cl
f)
CaC2 + H2O
CaO + H2C2
H2C2 + 2 [Cu(NH3)2]Cl
Cu2C2 + 2 NH4Cl + 2 NH3
g)
I2 + SO2 + 2 H2O
h)
–
I3
2 HI + H2SO4
i)
Pyridine acts as a base and thus shifts the equilibrium towards the products.
j)
Average consumption: Oil A = 1.62 mL, oil B = 1.44 mL
ions generate the brown colour. The use of starch solution is not possible because it has to be worked
aprotic (free of water) in this reaction.
Water content in % =
Water content of oil A =
Water content of oil B =
k)
–
0.09
100
Q = n(e ) · F
∙ 100 % = m consumption
1.44 mL ·4.8 mg/mL
∙ 100 %
∙ 100 % = 0.078 %
10000 mg
10000 mg
·t
weighed portion
1.62 mL · 4.8 mg/mL
1 mol H2O ≙ 1 mol I2 ≙2 mol e
m(H2O) =
V
mwater
mweighed portion
P 100 % = 0.069 %
–
-3
· 10 g = 9 · 10 g
n(H2O) =
9 ·10−3 g
18 g/mol
= 5 · 10 g ≙ 1 · 10 mol e
-4
-3
–
Q = 1 · 10 mol · 96485 C/mol  96.5 C
-3
Solution to problem 4-02
a)
b)
x = (2·0.926 V + 1.154 V):3 = 1.002 V
3+
–
(1) Au + 3 e
 Au
–
–
–
(2) [AuCl4] + 3 e  Au + 4 Cl
E1° = + 1.50 V
E2° = + 1.00 V
ΔG1°= -3∙F∙1.50 V
ΔG2°= -3∙F∙1.00 V
We have to find the equilibrium (= complex formation) constant Kco for the reaction
72
Answers Round 4 (theoretical)
3+

 [AuCl4]–
ΔG3 = -3∙F∙(1.50 V – 1.00 V)
ln Kco = 58.42
–
(3)
Au + 4 Cl
(3) = (1) – (2)
ΔG3 = - R∙T∙ln Kco
c)
1,20 V

ClO3
–
–
ClO4
1,18 V

ClO2
–
y
–
25
–
1,63 V
1,36 V

Cl2 
Cl
 3 ClO4– + Cl–
–
Disproportionation: 4 ClO3
–
1,65 V

HClO
Kco =2.34∙10
–
–
–
E°(ClO4 / ClO3 ) = 1.20 V
E°(ClO3 /Cl ) = (2·1.18 + 2·1.65 + 1.63 + 1.36) V / 6 = 1.44 V
y = (1.44 – 1.20) V = 0.24 V > 0
 G° = -n·F·y < 0
 The reaction disproportinates spontaneously.
d)

e)
+
It has to be
8,314·298
96485
E(H2O2, H /H2O) = 1.76 V +
+
V· ln[c(H )/c°]> – 0.15 V
5 H2O2 + 2 KMnO4 + 6 H
+
R·T
2·F
+
o 2
o
· ln[(c(H )/c ) ] > 1.61 V
+
+
ln[c(H )/c°] > -5.84
 5 O2 + 2 Mn + 8 H2O + 2 K
2+
(with c = 1 mol/L)
–3
c(H ) > 2.91·10
pH < 2.54
+
Solution to problem 4-03
a)
Two examples from the following table.
Expected
[Ar]3d44s2
[Ar]3d94s2
[Kr]4d35s2
[Kr]4d45s2
[Kr]4d65s2
[Kr]4d75s2
[Kr]4d85s2
[Kr]4d95s2
1 2
[Xe]4f 6s
2
[Xe]4f26s
Cr
Cu
Nb
Mo
Ru
Rh
Pd
Ag
La
Ce
Observed
5 1
[Ar]3d 4s
10 1
[Ar]3d 4s
4 1
[Kr]4d 5s
5 1
[Kr]4d 5s
7 1
[Kr]4d 5s
8 1
[Kr]4d 5s
10
[Kr]4d
10 1
[Kr]4d 5s
1 2
[Xe]5d 6s
1
1 2
[Xe]4f 5d 6s
Gd
Pt
Au
Ac
Th
Pa
U
Np
Cm
Expected
2
[Xe]4f86s
14
[Xe]4f 5d86s2
14
[Xe]4f 5d96s2
1 2
[Rn]5f 7s
2 2
[Rn]5f 7s
3 2
[Rn]5f 7s
2
[Rn]5f47s
5 2
[Rn]5f 7s
2
[Rn]5f87s
10
14
Observed
7
1 2
[Xe]4f 5d 6s
14
9 1
[Xe]4f 5d 6s
14
10 1
[Xe]4f 5d 6s
1 2
[Rn]6d 7s
2 2
[Rn]6d 7s
2
1 2
[Rn]5f 6d 7s
3
1 2
[Rn]5f 6d 7s
4
1 2
[Rn]5f 6d 7s
7
1 2
[Rn]5f 6d 7s
2
6
b)
Especially favoured are totally occupied (noble gas configuration, d , f , but also s , p ) as well as half
5 7
3
occupied (d , f , but also p ) electron shells.
c)
i) Fe : [Ar]3d
2+
14
10 2
vi) Pb : [Xe]4f 5d 6s
3+
5
3+
4
ii) Mn : [Ar]3d
3+
14
8
vii) Au : [Xe]4f 5d
4+
6
iii) Pd : [Ar]4d
2+
7
viii) Co : [Ar]3d
3+
3
2+
6
iv) Cr : [Ar]3d v) Fe : [Ar]3d
+
10
2+
2
ix) Cu : [Ar]3d x) Ti : [Ar]3d
d)
e)
73
Answers Round 4 (theoretical)
The degenerate orbitals in an octahedral ligand field split off into three levels of lower and two levels of
higher energy. The energy of the d-orbitals which lie between the axes of the coordinate system (dxy, dyz,
dyz) is lowered, that of the d-orbitals on these axes is lifted because of the stronger interaction with electrons of the ligands.
Energy
f)
g)
h)
4
d to d
7
3+
3+
Pd
4+
Fe
Mn
high-spin
5
4
4
low-spin
1
2
0
Cr
3+
2+
2+
Au
3+
Co
2+
Cu
+
2+
Fe
Pb
Ti
3
4
0
2
3
0
2
3
0
0
2
1
0
2
Solution to problem 4-04
a)
p(A2 )/p°
(p(A)/p°)2
= 2.1·10
3
3
-4
-3
p(A)/p(A2) = 8.9·10 bar/1.66·10 bar
b)
-4 2
-3
p(A2) = 2.1·10 · (8.9·10 ) bar
p(A2) = 1.66·10 bar
p(A)/p(A2) =0.54
k1 · c(A)· c(B) = k–1 · c(AB) 
Steady state of the equilibrium:
c(AB) =
k1
k−1
· c(A)· c(B)
v = keff · c(A)· c(B) = k2 · c(AB)
keff · c(A)· c(B) = k2 ·
k1
· c(A)· c(B)
k−1
keff = k2 ·
k1
-1
-1
-1
-1
keff = 25 s · 15 (mol/L) = 375 s (mol/L)
k−1
c)
The formation of the intermediate and its reaction back to the reactant have to be much faster than the
reaction of the intermediate to the product.
d)
G° = H° - T · S°
-1 -1
G° = 28.8 kJ/mol – 298 K · 97 Jmol K = -106 Jmol
G° = - R·T·ln K  ln K = 106 Jmol /(R·T)
-1
e)
G = G° + R·T·ln Q =
K = 1.04
c(CO2 ) c2(NH3 )
·
𝑐0
(𝑐0 )2
G° + R·T·ln c(NH2 )2CO c(H2 O)
·
𝑐0
𝑐0
2
0.097 ·0.02
G = -106 J/mol + R·T·ln
f)
ln K = 0.0428
Es ist G = G° + R·T·ln
0
(c = 1 mol/L)
G = -34.8 kJ/mol
0.85 ·1000/18
c(F)/c°
c(D)/c° · c(E)/c°
(1)
As reaction and back reaction are elementary in the equilibrium (eq) it follows that
k3 · ceq(D) · ceq (E) = k–3 · ceq (F)
K=
ceq (F)/c°
ceq (D)/c° · ceq (E)/c°
G° = - R·T·ln K
74
k–3
k3

k3
k−3
K=
=
ceq (F)
ceq (D) ·ceq (E)
k3
k−3
· c°
inserted in (1):
G = - R·T·ln K + R·T · ln
G = R·T·(ln

·
c(F)/c°
c(D)/c° · c(E)/c°
c(F)
)
c(D)·c(E)
G = - R·T · ln
G = R·T·ln
c(F)/c°
k3
· c° + R·T·ln
k−3
c(D)/c° · c(E)/c°
v3
v−3
-1
Answers Round 4 (theoretical)
g)
vob
v3
v−3
v3
=
=
v3 − v−3
v3
=1–
k−3 ·c(F)
v−3
v3
=
k3 · c(D)·c(E )
k−3
k3
v−3

= 0.5
· 0.4
v3
1.8 mol/L
mol/L· 0.9 mol/L
= 0.5
= 0.5
k−3

k3 ·co
= 0.1
k3
k−3
o
· c = K = 10
Solution to problem 4-05
a)
M(A)
w(A in AB3) =
M(A) + 3 ·M(B)
w(B in CB2) =
M(C) + 2 ·M(B)
w(C in CA) =
2 · M(B)
M(C)
M(C) + M(A)
= 0.2381

M(A) = 0.9375 · M(B)
= 0.7314

M(C) = 0.7345 · M(B)
=x

x=
0.7345 · M(B)
= 0.4393
0.7345 · M(B) + 0.9375 · M(B)
The mass percentage of C in CA is 43.93 %.
b)
On account of the empirical formulae only monovalent and divalent nonmetals have to be considered for
B: Hydrogen, halogens and chalcogens.
The solution can be found in a process of elimination by inserting the respective molar masses.
You find for B = bromine:
M(B) = M(Br) = 79.90
M(C) = 0.7345 · M(B) = 58.68
 Element C is nickel (M(Ni) = 58.69)
M(A) = 0.9375 · M(B) = 74.91
 Element A is arsenic (M(As) = 74.92)
AB3 = AsBr3 (arsenic tribromide) CA = NiAs (nickel arsenide, a semiconductor) CB 2 = NiBr2 (nickel bromide)
Remark: Neither hydrogen nor fluorine, chlorine and an element of the chalcogen group give reasonable
solutions. Only iodine provides elements and compounds if you allow small deviations:
You find for B = iodine:
M(B) = M(I) = 126.91
M(C) = 0.7345 · M(B) = 93.22
 Element C could be niobium (M(Nb) = 92.91)
M(A) = 0.9375 · M(B) = 118.98
 Element A could be tin (M(Sn) = 118.71)
 CA = NbSn, AB3 = SnI3 and CB2 NbI2. None of these compounds is known.
Solution to problem 4-06
a)
Octant
1
III
1
II
Unit cell (8 x octant)
III
4 x /8 Fe
b)
 Empirical formula:
4 Fe
II
3 x /8 Fe
3 Fe
3x¼O+1xO
14 O (H2O)
9 x ¼ CN
18 CN
III
K[Fe Fe (CN)6] + 2 K
3+
II
K[Fe Fe (CN)6] + 3 K
Fe + K4[Fe (CN)6]
2+
III
3–
c)
Fe + [Fe (CN)6]
d)
Fe + K4[Fe (CN)6]
2+
II
II
III
II
Fe 4Fe 3(CN)18 · 14 H2O
2+
Fe + K3[Fe (CN)6]
III
Fe 4Fe 3C18N18H28O14 or
III
II
+
III
II
+
3+
II
4–
Fe + [Fe (CN)6]
II
II
+
K2[Fe Fe (CN)6] + 2 K
75
Answers Round 4 (theoretical)
The reaction product contains only iron(II) and thus an electron transfer between iron ions of different
oxidation states via the cyanide ions, which causes the color, is not possible.
e)
Coordination spheres:
Iro(II) is surrounded by six cyano groups:
4–
formal [Fe(CN)6]
3–
one iron(III) is surrounded by six cyano groups: formal [Fe(NC)6]
three iron(III) are each surrounded of four cyano
–
groups and two water molecules :
formal [Fe(H2O)2(NC)4]
f)
Diluted solution:
half-concentrated hydrochloric acid:
concentrated hydrochloric acid:
2+
[Cu(H2O)6]
[Cu(H2O)4Cl2]
2–
[CuCl4]
–
(or [Cu(H2O)3Cl3] )
2–
(or [CuCl4(H2O)2] )
g)
O H
O
H O
B
”
B
O
O
O
”
B
B
O H
O
O H
h)
i)
j)
2–
The other anions are disulfide anions (S2 ). Cu2S · CuS2 (= 3 CuS).
4 HCl + O2
2 Cl2 + 2 H2O
CuCl2 + ½ O2
CuO + Cl2
CuO + 2 HCl
CuCl2 + H2O
Solution to problem 4-07
a)
b)
c)
d)
e)
f)
g)
h)
-1 -1
W = - 2 atm · 101.3 kPa/atm (
i)
U = H = 0
j)
Q = 3.93 kJ
k)
lg 3 = 3.99 -
76
-3
3
3
p = n · R · T/V p = 2 mol · 8.314 Jmol K · 273 K /11.2·10 m
p = 405.3 · 10 Pa
Cp: Heat capacity at constant pressure
CV: Heat capacity at constant volume
Cp > CV because when heating at constant pressure work is done by the system. Since work is done in the
case of Cp it follows that for a given amount of heating the temperature rise is less and therefore that Cp
-1 -1
is greater.
Cp - CV = R

Cp = 29.4 Jmol K
Zero
-1 -1
U = n · CV · T
U = 2 mol · 21.1 Jmol K · 100 K U = 4.22 kJ
4.22 kJ
-1 -1
-3 3
3
p = n · R · T/V p = 2 mol · 8.314 Jmol K · 373 K /11.2·10 m p
p = 553.8 · 10 Pa
-3
3
3
3
U = H – p · V U = H - (p · V) H = 4.22 kJ + 11.2 · 10 m (553.8 · 10 Pa - 405.3 · 10 Pa)
3
3
H = 4.22 · 10 J + 11.2 m · 148.5 Pa
H = 5.899 kJ
-1 -1
(or H = n · Cp · T H = 2 mol · 29.4 Jmol K · 100 K
H = 5.88 kJ)
Work done = p,V-work W = - p · V
443
𝑇𝑆 /K − 0,49
2 mol ·R ·373 K
2 ·101,3 kPa
 TS/K = 0,49 +
-3
3
– 11.2 · 10 m )
443
3,99 −lg 3
W = -3.93 kJ
TS = 126.6 K = -146.6 °C
Answers Round 4 (theoretical)
l)
U = H - (p · V)
 H = U + (p · V)
V(CH4(l)) ≪ V(CH4(g))
 V(CH4(l))  0 L
 with p · V = n · R · T for 1 mol CH4:
(p · V) = R · 112 K – 0 J/mol = 931 J/mol
H = U + (p · V)
H = 7.25 kJ/mol + 0.931 kJ/mol  8.18 kJ/mol
32.5 kJ are removed
 32.5/(8.18 kJ/mol) = 3.97 mol CH4 must evaporate
that are V =
3,97 mol ·R ·112 K
101300 Pa
-3
3
= 36.5·10 m = 36.5 L CH4
Solution to problem 4-08
a)
b)
A
B
C
D
E
F
G
(–)-Muscone
X = MgBr2
2
Et2O
MgBr
MgBr2
Mg
c)
d)
Y = CrO3
77
Answers Round 4 (theoretical)
e)
Z: Ethene
Solution to problem 4-09
a) In cyclohexane the C-C bond angles are very near to 109.5°, the ideal tetrahedral value. In Cyclopropane these angles are with 60° clearly smaller.
The C-C bond length in cyclopropane is a bit smaller than in cyclohexane.
The better the overlapping of the orbitals the stronger the bond. The
eclipsed position of the hydrogen atoms in cyclopropane weakens the bond
compared with the not eclipsed position in cyclohexane.
 The C-C bonds in cyclopropane are weaker than in cyclohexane.
H
b)
H
H
H
H
H
Mechanism of the reaction:
secondary carbenium
ion (more stable)
Mainly the more stable secondary carbenium ion is formed and consequently 2bromobutane is formed. The observed regioselectivity follows the rule of Markovnikov.
primary carbenium
ion (less stable)
c)
Energy
Transition state I
Intermediate: carbenium ion
Transition state II
Reactants
Products
Reaction coordinate
78
d)
Hybridization state of the carbon
2
atom in the center: sp
Answers Round 4 (theoretical)
e)
Simmons-Smith reaction of (Z)-1,2-Diphenylethene (Remark: The reaction equation is not asked for):
(1R,2S)-1,2-Diphenyl cyclopropane
resp.
(1R,2S)-1,2-Diphenyl cyclopropane
f)
Br
OH
1
2
3
4
g)
Remark: The methylene hydrogen atoms in the cyclopropane rings are chemically equivalent.
Solution to problem 4-10
a)
b)
c)
d)
A
B
C
E
F
G
D
i):
Oxidation with K2Cr2O7, dil. H2SO4 (Jones Oxidation)
or with CrO3, dil. H2SO4, acetone
or with KMnO4
+
ii): Acidic (H3O ) or basic (OH ) opening of the epoxide
iii): Periodate cleavage with NaIO4
e) MeLi would only react violently accompanied by deprotonation of the
+
acid.
f)
Ethane diol, H
g)
Hemiacetal
Acetal
79
20. – 29. July 2015
Examinations
80
IChO Baku: Theoretical Test
Theoretical Test
Physical Constants, Units, Formulas and Equations
Universal gas constant
R = 8.3145 J∙K–1∙mol–1
Standard pressure
p = 1 bar = 105 Pa = 750 mmHg
Atmospheric pressure
1 atm = 1.013105 Pa = 760 mmHg
Zero of the Celsius scale
273.15 K
Reversible adiabatic process for an ideal gas
pV 1 R /CV = const
Work made by an ideal gas in an adiabatic process
W = nCV (T1 – T2)
Dependence of internal energy on temperature
U(T2) = U(T1) + CV (T2 – T1)
Gibbs energy
G = H – TS
Relation between equilibrium constant and standard
Gibbs energy
 G
K = exp  
 RT

Dependence of Gibbs energy of reaction on concentration or pressure
Change of Gibbs energy in time for the system with
two chemical reactions




G = G  RT ln
aprod
areag
,
a = c / (1 mol/L) for the substances in solution, a = p / (1 bar) for gases
GSyst
t
 G1r1  G2 r2
81
Problem 1. New and well-forgotten old refrigerants
The problem of choosing a refrigerant for refrigeration and air conditioning systems attracted the
attention of scientists and technologists throughout the last century. It has been suggested that during this time refrigerants progressed through four generations. Ammonia, which was ascribed to the
first generation, had been used in most of the oldest refrigeration units. It was later replaced by
chlorofluorocarbons (CFCs) – derivatives of methane and ethane
with the hydrogen atoms replaced by fluorine and chlorine.
In Baku, at "Bakkonditsioner" factory, production of the first Soviet serial household air conditioners BK-1500 had been
launched. A second-generation refrigerant chlorodifluoromethane CHF2Cl was used in them. In this problem, we compare
various refrigerants in terms of thermodynamics.
First air conditioner of Baku factory in a souvenir shop in the Old City
(“Icheri Sheher”)
Thermodynamic properties of various refrigerants
Refrigerant
NH3
CHF2Cl
CF3CH2F
CF3CF=CH2
ΔHvap / kJ·mol–1(at 280 K)
21.3
20.0
22.1
19.1
“Generation”
1
2
3
4
Cv (gas) /J·K–1·mol–1
26.7
48.8
79
120
Consider a model refrigeration cycle consisting of 4 steps schematically shown below in the pressure
(p) – internal energy (U) coordinates.
p
p2
3
liquid
p1
2 (T2)
liquid + gas
gas
0 (T1)
1 (T1)
U
During the first step of the cycle (line 0-1 in diagram), a liquid refrigerant is boiling at constant pressure p1 and temperature T1 (boiling temperature) until it completely evaporates. At this step, the
refrigeration unit absorbs heat from surrounding objects. At the second step, the refrigerant undergoes reversible adiabatic compression and heats up to temperature T2 (line 1-2). After that the compressed refrigerant is cooled in a condenser at constant pressure p2 (line 2-3) and then returns to the
initial state (line 3-0).
82
IChO Theoretical Test
Let the cycle involve 1 mole of refrigerant, which is initially (point 0) completely liquid, T1 = 280 К, T2
= 380 К, assume that the vapor of any refrigerant behaves like an ideal gas. The thermodynamic
characteristics of refrigerants are listed in the table above.
1.1. For each of refrigerants, ammonia and chlorodifluoromethane, calculate the amount of heat Q
absorbed by refrigeration unit during heat exchange (line 0-1) and the work W required to compress its vapor (line 1-2).
1.2. Which quantity(ies) (U, H, S, G, V) remains constant during the compression step? Indicate it by
the circle.
To compare the energy efficiency of refrigeration cycles with different parameters and refrigerants,
the coefficient of performance (COP) is used, which is defined as a ratio of heat removed from a
cooled system to the work of compressor: COP = Q/W.
1.3. Calculate the values of COP in a considered cycle for ammonia and chlorodifluoromethane.
2.1. Why was ammonia replaced by CFCs in household refrigeration units? (Choose only one option)
a) to increase the energy efficiency of refrigeration cycles
b) to reduce the cost of refrigerant
c) for environmental care
d) for user security reasons
A search for replacement of CFCs as refrigerants started when it was shown that their use can cause
irreparable damage to the protective ozone layer of the atmosphere. The third, ozone-friendly generation of refrigerants came on the scene. Its typical representatives are fluoroalkanes.
2.2. What is the cause of the damage made by CFCs to the ozone layer? (Choose only one option)
a) ozone molecule easily adds to C–F bond
b) C–F bond is easily broken by radiation, which leads to the formation of free radicals
c) ozone molecule easily adds to C–Cl bond
d) C–Cl bond is easily broken by radiation, which leads to the formation of free radicals
However, under the 1997 Kyoto Protocol, fluoroalkanes also have to be replaced because they accumulate in the atmosphere and rapidly absorb infrared radiation, causing a rise in temperature of
the atmosphere (the greenhouse effect). The refrigerants of the fourth generation such as 2,3,3,3tetrafluoropropene CF3CF=CH2 have been suggested and are coming into use.
2.3. Why does this compound enhance the greenhouse effect less than fluoroalkanes? (Choose only
one option)
a) it absorbs less infrared radiation
b) it is more reactive and easier to decompose
c) it easily reacts with ozone
d) it is better soluble in water
3.
Calculate the values of the COP in the refrigeration cycle considered above for two refrigerants
of the third and fourth generations – CF3CH2F and CF3CF=CH2. Did the energy efficiency improve
in comparison with CF2Cl2? Choose “Yes” or “No”.
83
Unlike household appliances, industrial refrigeration systems are often still using ammonia. It does
not contribute to the greenhouse effect nor does it destroy the ozone layer. Industrial units can have
a huge size and a large cost. Prior to their construction, they should be carefully modeled taking into
account many different factors. In real systems, some part of the refrigerant at the start of the heat
exchange with the environment is in the vapor phase (point 0 in the diagram below), and at the end
(point 1) it is always overheated above the boiling point.
p
3 (T3)
p2
liquid
p1
liquid +
gas
0 (T0)
2 (T2)
gas
1 (T1)
U
Consider a cycle with 1 mole of ammonia. Its thermodynamic properties are the following: enthalpy
of vaporization ΔHvap = 23.35 kJ·mol–1 at Tvap = 239.8 К (boiling temperature at 1 bar pressure). Heat
capacity of the liquid phase Cv(liq) = 77 J·K–1·mol–1, of the gas phase Cv(gas) = 26.7 J·K–1·mol–1. Assume
that the heat capacities are temperature-independent and the vapor behaves like an ideal gas. The
temperature dependence of the saturated vapor pressure of ammonia can be described by the empirical equation:
log (p/bar) = 4.87 – 1114 / (T/K – 10.4).
During the first step of the cycle (line 0-1 in diagram), the equilibrium mixture of liquid refrigerant
and its vapor receives heat from the environment at constant pressure p1 = 3.0 bar. The refrigerant
completely evaporates and overheats up to the temperature T1 = 275 K. In the beginning of the process (point 0), the molar fraction of gaseous ammonia is x = 0.13.
4.1. Calculate the initial temperature of refrigerant T0, its volume change ΔV and the amount of heat
Q absorbed by refrigeration unit during this step.
Then the refrigerant is reversibly and adiabatically compressed. It heats up to the temperature T2 =
393 К (line 1-2).
4.2. Find the work W required for compression and the COP of the system.
At the next step corresponding to the line 2-3 in diagram, the compressed refrigerant is cooled in a
condenser at constant pressure. Then it returns to the initial state through adiabatic expansion with
zero work (line 3-0).
4.3. Determine the temperature T3 at point 3 to which the refrigerant is cooled in a condenser.
In the production of refrigeration units it is necessary to consider climatic factors. If a condenser is
cooled by atmospheric air, the temperature T3 is higher as higher is the air temperature.
4.4. How will the COP change if T3 increases while T0, T1, T2 remain the same?
a) Increase
b) Remain the same
c) Decrease
84
IChO Theoretical Test
Problem 2. Coupling of chemical reactions
I.Prigogine (left)
N. Shilov
W. Ostwald
When in the system one reaction allows another one to proceed they say that these two reactions
are coupled. Ilya Prigogine, Nobel prize winner in chemistry (1977) in his books widely used the concept of “coupled reactions”. Coupling of reactions is an essential feature of living systems, including
human body.
How one reaction makes another one to occur? In this problem we are going to discuss several possible mechanisms of coupling.
(I) “Chemical coupling”
“On Chemical coupling” was the title of the dissertation defended by Russian chemist N.Shilov in
1905. N. Shilov was the graduate student of famous professor W. Ostwald from Germany. Dr. Shilov
described the following set of reactions.
The substance А does not react with Ac. In the presence of the third reagent (called inductor), In,
however, the reaction of А with Ac takes place:
In the absence of In
A  Aс 
 no reaction!
(1)
A  Ас 
 Р1
(2)
In the presence of In
In is not a catalyst! Its concentration decreases in the course of the reactions.
According to the mechanism proposed by Shilov, Ас reacts not with A itself, but with the intermediate product R of the reaction of А with In:
(a )
A  In 
 R 
 P2
( b)
R  Ас 
 P1
(3)
 and  are stoichiometric coefficients. Other stoichiometric coefficients and all partial reaction orders in both reactions are equal to unity.
In the Shilov’s experiments the ratio of the consumed amounts of Аc and In, I 
n Ас
increased up
nIn
to the constant value with the increasing initial concentration [Ac]0 at [In]0 = const.
1.1. What was this limiting constant value of I at [Ac]0  , [In]0 = const?
1.2. Plot on the graph the dependence of I vs [In]0 at [Ac]0 = const. Assume that In was completely
consumed and Аc was in excess. Use the steady-state approximation if necessary.
85
What if Shilov’s mechanism is not valid and In is a conventional catalyst of the reaction (2)? Simultaneously In reacts with А and its concentration decreases. The reaction mechanism in this case is
(a )
A  In 
 P2
( b)
In, catalysis
A  Ас  P1
(4)
1.3. What is the limiting value of I for the reaction mechanism (4) at [Ac]0  , [In]0 = const?
(II) «Kinetic coupling»
The Gibbs energy of the gas-phase reaction
k5

 HBr  H
Br  H 2 

k5
(5)
is positive, G(5) = 66 kJmol–1 at Т = 600 К and standard pressures of all reactants and products.
2.1. What is the ratio of the rates of forward and reverse reactions,
r5
, under such conditions?
r5
Reaction (5) proceeds in the forward direction due to the reaction (6) which simultaneously occurs in
the system:
k5

 HBr  H
Br  H 2 

k5
k6
H  Br2 
HBr  Br
(6)
k5, k–5, k6 are rate constants of forward and reverse reaction (5) and forward reaction (6), respectively.
This is the kinetic coupling of two reactions.
Let pressures of neutral molecules keep standard values p(H2) = p(Br2) = p(HBr) = 1 bar, and pressures of radicals p(H), p(Br) reach quasi-stationary values. Rate constant k6 is 10 times larger than k–5.
2.2. Calculate G(5) and
r5
under such conditions.
r5
2.3. Compare the results obtained in 2.1 and 2.2. How does kinetic coupling affect the reaction (5)?
(a) It makes forward reaction to proceed faster,
(b) It changes the sign of the observed rate robs = r5 – r–5 ,
(c) It makes reverse reaction to proceed slower.
Choose only one
(III) ”Second law of thermodynamics restricts coupling”
According to the Second Law of thermodynamics occurring of two simultaneous chemical reactions
in the system should decrease the system’s Gibbs energy Gsyst,
GSyst
t
 0.
One of these reactions may have positive Gibbs energy and still proceed in the forward direction due
to the coupling with the second reaction. This second reaction must have negative Gibbs energy and
the requirements of the Second law must be fulfilled! Consider the example.
The synthesis of urea
86
IChO Theoretical Test
2NH3 + CO2 = (NH2)2CO + H2O
(7) G(7) = 46.0 kJmol–1
is supposed to be coupled with the complete oxidation of glucose
1/6 C6H12O6 + O2 = CO2 + H2O
(8) G(8) = –481.2 kJmol–1, r(8) = 6.010–8 M–1min–
1
.
Both reactions are presented schematically. No other reactions are considered.
3.1. What is the maximum rate of the reaction (7) permitted by the Second Law?
Assume that partial pressure of CO2 in the system was twice increased.
3.2. By how many times will the maximum permitted rate of the reaction (7) be changed under such
conditions? Temperature is 310 К.
Problem 3. Two binding centers – competition or cooperation?
Many chemical reactions in living organisms include the formation of “host-guest” complexes where
the host molecule reversibly binds one or several guest molecules. Consider a host molecule H with
two binding centers – say, a and b which have different affinity to guest molecules:

 HGa
H+G 

Ka =
[ HGa ]
[ H ][G ]

 HGb
H+G 

Kb =
[ HGb ]
[ H ][G ]
Kb  Ka.
where Ka,b are the binding constants for the centers a and b, brackets denote molar concentrations.
Attachment of one G molecule to H can change the binding ability of the second centre. This change
is described by the “interaction factor”  which reflects the influence of one binding center on another and is defined as follows:

 HG2
HGa + G 

[ HG2 ]
= K b
[ HGa ][G ]
where HG2 is the completely bound complex.
1.1. Determine the range of values (or one value, if necessary) of  which correspond to three possible ways of interaction between binding centers: a) cooperation (binding by one center facilitates subsequent binding); b) competition (first binding complicates the second); c) independence (no interaction).

 HG2.
1.2. Find the equilibrium constant for the process: HGb + G 

The solution was prepared with the initial concentrations [H]0 = 1 M and [G]0 = 2 M. After the reactions were completed, the concentration of H decreased by 10 times and that of G – by 4 times. For
these host and guest, Kb = 2Ka.
2.1. Determine the concentrations of all other species in the solution and find the binding constant Ka
and the factor .
87
2.2. Determine the relative Gibbs energies of H and all host-guest complexes2 at [G] = 1 M and  = 3.
In the scheme on the answer sheet, write the corresponding formula near every line.
Some amount of G was added to 1 mole of H and the mixture was dissolved in water to obtain 1 liter
of the solution. The number of the totally bound molecules HG2 in the solution is equal to the total
number of single-bound molecules HG.
2.3. Find the initial amount of G (in mol). The constants Ka and Kb and the factor  are the same as in
question 2.1.
2.4. What would be the equilibrium composition of the solution if: a)  = 0; b)  is very large (  ).
The constants Ka and Kb as well as the initial concentrations of H and G are the same as in question 2.1.
Problem 4. From one yellow powder to another: A simple inorganic riddle
The yellow binary compound X1 was completely dissolved in concentrated nitric acid by heating, the
gas evolved is 1.586 times denser than air. Upon adding an excess of barium chloride to the solution
formed a white solid X2 precipitates. The filtrate reacts with an excess of silver sulfate solution forming a precipitate of two solids X2 and X3, also separated by filtration. To the new filtrate the solution
of sodium hydroxide was being added drop-wise until the solution became nearly neutral. At this
time a yellow powder X4 crystallized from the solution. The mass of X4 is nearly 2.4 times larger than
that of X2.
1.
Determine the chemical formulae of X1 – X4.
2.
Write down all reaction equations in molecular form.
3.
In the structural unit of X1 all atoms of one element are in equivalent positions. Draw the structure of X1.
Predict the products of X1 interaction with:
a) excess oxygen;
b) excess concentrated sulfuric acid;
c) solid KClO3 at friction.
Write down the reaction equations.
4.
Problem 5. Indispensable glucose
Carbohydrates are the most important providers of energy for living cell. Monosaccharide glucose is
a source of energy for the living cell, but for persons who suffer from diabetes glucose may be dangerous. High level of glucose may lead to hyperglycemia and even death. That is why people avoid
consuming too much carbohydrates and glucose particularly.
1. Determination of reducing sugars in fruit juice
One of the technique for determination of reducing sugars in different samples includes the use of
Fehling's reagent. This reagent was prepared by mixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassium-sodium tartrate and sodium hydroxide (solution B). A 10.00-mL aliquot of fruit
2
Assume for simplicity that G(G) = 0
88
IChO Theoretical Test
juice was transferred into a titration flask and Fehling's reagent was added. Solution C thus obtained,
was then heated and red precipitate was formed.
1.1. Write the balanced ionic equation of chemical reaction occurring upon heating of the solution C.
Use Cu2+ for copper.
After that 10 mL of 10% solution of potassium iodide and 1 M sulfuric acid were added to the flask.
The mixture was covered with watch glass and was then placed in a dark place. An excess of iodine
was then titrated with 0.05078 М sodium thiosulphate solution. 11.87 mL of the titrant was required
to reach the endpoint.
1.2. What happened within the flask? Write the balanced equation(s) in molecular or ionic form.
1.3. Assuming the initial sample contained only glucose and fructose, and all fructose was transformed into glucose under the experimental conditions, calculate the total mass content of sugars (in g/L) in a fruit juice.
A new 10.00-mL aliquot of the same juice was treated with a 10.00-mL portion of acidified potassium iodate solution (0.01502 М) and 10 mL of 10 % solution of potassium iodide. After the mixture
has darkened, an excess of sodium hydroxide solution was added. The flask was then covered with a
watch glass and put into a dark place. The obtained solution was acidified and titrated with 0.1089
M solution of sodium thiosulphate. The average titrant volume used for titration was 23.43 mL.
1.4. Write the balanced equation(s) for the described reactions in molecular or ionic form.
1.5. Why is it better to synthesize iodine in situ (in the same titration flask) than to add its standard
solution? Choose one answer.
a) The aqueous iodine solution is unstable
b) Iodine evaporates
c) Iodine precipitates
d) Standard solution of iodine cannot be obtained
1.6. Calculate the mass content of each sugar (in g/L) in the juice.
1.7. One bread exchange unit (1 BEU) corresponds to the content of 12 g of digestible carbohydrates
in product. How many BEU are in one glass (200 g) of juice?
2. Diagnosis of diseases
The derivative of glucose, 2-deoxy-2-(18F)fluoro-D-glucose (FDG), is the most common radiopharmaceuticals for diagnosis of cancer using positron emission tomography. The first step of FDG preparation is to produce a radionuclide fluoro-18 by nuclear reaction in a cyclotron. The next step is the
radiochemical synthesis. Fluorine-18 is introduced into D-glucose molecule by nucleophilic substitution. 18F-2-deoxy-D-glucose once injected into the patient actively accumulates in cells of malignant
tumors; this process is accompanied by decomposition of fluorine-18. This radionuclide is a β+ emitter – nucleus emits a positron (anti-electron). Positron interacts with an electron and after that annihilation occurs, which can be detected. This allows determining precisely the tumor location.
89
2.1. Complete the nuclear reactions leading to various fluorine isotopes.
a) 18O + p  n + …
b)
… + 12 D  18F + 
c)
19
b)
16
F + 12 D  20F + …
O + …  18F + p + n
The decay mode of unstable light nuclei depends on the ratio between neutrons and positrons in
them. If this ratio is greater than that for a stable isotope then the nucleus decays in a β–-mode, if it
is smaller – in a β+-mode.
2.2. Determine the type of decay for the following nuclei: 11С, 20F, 17F and 14C.
When nuclear reaction (a) is used for fluorine-18 preparation, the target material is presented as
water enriched with H218O. The presence of usual water H216O leads to a side nuclear reaction with
16
O, leading to the formation of isotope 17F.
2.3. It is known that within five minutes after completion of irradiation the target the ratio of radioactivities of 18F and 17F is 105. Assuming that the radioactivity of each isotope is proportional to
the nuclear reaction yield and the mole fraction of a component in the irradiated target, calculate the mass fraction of H218O in the target. T1/2(18F) = 109.7 minutes, T1/2(17F) – 65 seconds. The
ratio between nuclear reactions yields is η(18O→18F) / η(16O→17F) = 144.7.
2.4. Calculate the yield of labeling D-glucose with fluorine-18, if initial radioactivity of a fluorine-18
sample was 600.0 MBq and radioactivity of the obtained 18F-2-deoxy-D-glucose is 528.2 МBq.
Synthesis time is 3.5 minutes.
2.5. Biological half-life of 18F-2-deoxy-D-glucose is 120.0 minutes. How much radioactivity (in MBq)
will remain in the patient one hour after injection of FDG with the initial radioactivity of 450
MBq.
Problem 6. Bread is the staff of life
When you pass by the bakery, you are stopped by the smell of freshly baked bread. The hero of one
of the novels said on a similar occasion: "If you tell me that this is not perfect, you are my enemy
forever." The principal bread flavor component was identified in 1969 as compound X which occurs
in equilibrium with its tautomer Y in a 2:1 ratio. Unfortunately, both forms are labile, and after some
hours bread has no the same nice smell.
This tautomeric mixture of X and Y was synthesized in 1993 from piperidine by the reaction sequence given in Scheme 1.
Scheme 1.
It is noteworthy that the initial ratio of X and Y was 1:4; on standing this ratio gradually changed to
the equilibrium one.
90
IChO Theoretical Test
Compound A which is characterized by 3-fold axis of symmetry (i.e., rotation by 120 results in a
molecule indistinguishable from the original) occurs in equilibrium with its diastereomer B. The interconversion of these two forms proceeds via very reactive intermediate E which is also intermediate in A and B formation as well as their transformation to C. Compounds A, B, and E have the same
elemental composition: C = 72.24%, H = 10.91%, N = 16.85%.
1. Write down the structural formulae of compounds A-E, X, Y.
Treatment of compound D with MeLiLiBr complex in Et2O at 0 C failed to produce the target products X and Y. Instead, a yellow precipitate F was initially formed. Aqueous workup of this precipitate
led to the mixture of compound D and its tautomer G.
2.
Write down the structural formulae of compounds F and G.
Another approach to compound X is based on the use of pipecolinic acid derivartive H. It was shown
that X can be synthesized by reaction sequence presented in Scheme 2.
Scheme 2.
3.
Write down the structural formulae of compounds I and J
Problem 7. Not by bread alone
Pomegranate is called in Azerbaijan, which is famous for its vegetables, as
the “king of all fruits”. Pomegranate is honored in various religions as a
“fruit of Paradise”, symbol of righteousness, wealth, hope for eternal life.
In 1878 alkaloid pelletierine was isolated from the bark of pomegranate
tree (Punica granatum L., Lythraceae). This alkaloid is traditionally used as
an anti-helminthic drug. Initially a wrong structure of (3-(piperidin-2yl)propionaldehyde) XW was proposed for pelletierine. But now it is accepted that natural pelletierine is (S)-1-(piperidin-2-yl)propan-2-one (XS).
1. Write down the structural formulae of XW and XS (the latter – with the stereochemical information).
The synthesis of natural pelletierine (XS) based on the transformation of nortropanol A was recently
described.
91
2.
Write down the structural formulae of compounds B-G with the stereochemical information.
3.
Nortropanol A was used in this reaction as a single stereoisomer. How many stereoisomers can
exist for compound A?
Enantiomer of XS was synthesized using chiral tert-butanesulfinamide (H):
4.
Write down the structural formulae of compounds I-L with the stereochemical information.
Problem 8. Oil for Life and Life after Oil
Azerbaijan is known for its vast oil and gas fields. The first drilling for oil
was done in Bibi-Heybat in 1846, 13 years before establishment of the
first commercial oil well in Pennsylvania (USA). This remarkable date in
the history of Azerbaijan is regarded as a starting point of contemporary
oil industry, the leading sector of today’s world economy. Currently, onland and shelf sea oil production is being developed in Azerbaijan.
Though serious precautions are taken, there is always a risk of hydrocarbon pollution of the environment during production, transportation, and processing of oil. In this
task we will consider diverse technologies of oil spills clean up and specific features of metabolic
pathways involved.
Application of complex solvents (dispersants) leading to capture of marine oil spills is among most
promising clean up approaches. Organic substance X (11.94% of H by mass) is a typical component of
such dispersants. Safety of X to human is fiercely debated. X1 (54.53% of carbon by mass) composed
92
IChO Theoretical Test
of three elements and excreted with urine is the major metabolite of X in humans. The numbers of
atoms of different elements in X1 are three consecutive terms of a geometric progression, whereas
the sum of these numbers does not exceed 25.
1a. Decide on the relationship (tick the correct variant) between the numbers of carbon and oxygen
atoms in X1
n(C) > n(O) 
n(C) < n(O) 
n(C) = n(O) 
Data insufficient 
1b. Derive the empirical formula of X1 (hereafter always show your work where required).
The biotransformation of X into X1 occurs in two enzymatically catalyzed steps according to the hereunder reaction equations (NAD+ and NADH are the oxidized and reduced forms of nicotinamide adenine
dinucleotide, respectively):
Х + NAD+ → X0 + NADH + H+
(1)
+
+
X0 + NAD + H2O → X1 + NADH + H
(2)
1c.
Derive the molecular formula of X.
A minor metabolic transformation of X is catalyzed by cytochrome P450-dependent monooxygenase. This reaction leads to two compounds X2 (51.56% of oxygen and 9.74% of hydrogen by mass)
and X3.
1d.
Derive the molecular formula of X2 and draw its structure.
X contains only primary and secondary carbon atoms. X0 and X3 contain common functional group.
1e. Draw the structural formulae of X, X1, and X3.
In a medical study, personnel working with X-based solvents without proper protection was found to
have a stationary concentration of X in blood.
2.
Choose the graph of X1 daily mass content in the body of a volunteer participated in this experiment. Write down the number of the correct graph.
2
3
0
6
12
18
time, h
24
mass
mass
mass
1
0
0
6
12
18
24
6
12
18
24
time, h
time, h
93
5
6
0
6
12
18
24
time, h
mass
mass
mass
4
0
6
12
time, h
18
24
0
6
12
18
24
time, h
The use of different bacteria is also considered as a promising way for the removal of hydrocarbon
(even aromatic) contaminants from sea water and soil. Under aerobic conditions, benzene undergoes biodegradation as follows (first three steps are balanced):
Under the same conditions, a monocyclic aromatic hydrocarbon P (91.25% of carbon by mass) undergoes the following transformation (first three steps are balanced):
P3 gives a positive iodoform test. A 100 mg sample of P3 requires 6.41 mL of 0.100 M KOH solution
for complete neutralization.
3.
Derive the structures of P–P3.
Microorganisms Alicycliphilus are capable of biodegradation of aromatic hydrocarbons even under
anaerobic conditions. The process requires a suitable electron acceptor such as inorganic anion Y1
(in soil).
The intermediate anion Y2 is enzymatically decomposed according to the reaction equation:
Y2(aq) = Y3(aq) + Y4(g),
wherein each of Y3 and Y4 is composed of atoms of only one element. T2 does not contain identical
functional groups, and by contrast to Y3 gives a precipitate when treated with the ammonia solution
of Ag2O.
4.
94
Deduce and draw the structures of Y1-Y4, T1-T2.
IChO Baku: Practical Test
PRACTICAL EXAMINATION
JULY 23, 2015
List of Chemicals
Concentration
Name
State
Quantity
3-Methylthiophene
1-Bromo-2,5-pyrrolidinedione (NBS)
Carbon tetrachloride
Solution in
CCl4
4g/8 mL
4g
Solid
-
7.3g
Liquid
-
24 mL
Unknown catalyst
in CCl4
Potassium carbonate
Solid
Labeled
Hazard Codes
Task 1
3-methylthiophene in CCl4
NBS
7,3 g
CCl4
H225, H302, H332
H302, H314
H301, H331, H311, H317,
H351, H372, H402, H412
Catalyst
-
0.02 g
K2CO3
H315, H319
Task 2
Test solution containing VO2+ and Cr3+
Sulfuric acid
Potassium permanganate
Oxalic acid
Phenylanthranilic
acid
Ammonium iron(II)
sulfate
Silver nitrate
Ammonium persulfate
Aqueous
solution
Aqueous
solution
Aqueous
solution
Aqueous
solution
Aqueous
solution
Aqueous
solution
Aqueous
solution
Aqueous
solution
To be deter-mined
100 mL
Test solution
H302, H312, H314, H332
1M
~ 500 mL
1M H2SO4
H290, H302, H314, H332,
H351
0.03 M
15 mL
0.03 M KMnO4
H272, H302, H400, H410
0.03 M
30 mL
0.03 M Н2С2О4
H314, H318
0.1 %
5 mL
Indicator
Read from
the label
100 mL
Mohr's salt
H302, H315, H319, H335
0.3 %
5 mL
0.3 % AgNO3
H272, H302, H314, H410
10 %
70 mL
10 % (NH4)2S2O8
H272, H302, H315, H317,
H319, H334, H335
Task 3
Diclofenac containing
medicine
Potassium permanganate
Sulfuric acid (in the
same bottle as for
Task 2)
Diclofenac sodium
salt
Aqueous
solution
Aqueous
solution
To be deter-mined
5 mL
Control
H301
610-3 M
~ 30 mL
KMnO4
6×10-3 M
H272, H302, H400, H410
Aqueous
solution
1M
~ 500 mL
1M H2SO4
H290, H302, H314, H332,
H351
Aqueous
solution
~ 600
mg/L
~ 20 mL
DCF
600 mg/L
H301
The hazard statements description was given.
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IChO Baku: Practical Test
List of labware and equipment
Item
Quantity
On the tables for common use
Refractometer Refracto 30GS
Napkins for refractometer cleaning
Wash bottle “Cleaning solvent” for the refractometer
Aluminum foil for wrapping
Balances
Gloves (S, M, L)
Large bottle labeled “H2O dist.”
Napkins for general purposes
1-2 / 1 lab
1-2 rolls / 1 lab
1-3/ 1 lab
1 Pack / 1 row
Item
Quantity
Located
Under the hood
Under the hood
Under the hood
On lab assistants’ table
On separate tables
On lab assistants’ table
Near the sink
Near the sink
As labeled on Fig. 1, 2, 5
On each working place, to be used in more than one task
Hot-plate magnetic stirrer
Waste bottle labeled “Waste”
Cotton gloves
Wash bottle, 500 mL, labeled “H2O distilled”
Pipette pump, 10 mL, green
Pipette pump, 2 mL, blue
Graduated cylinder, 25.0 mL for H2SO4 only
Safety goggles
Napkins for general purposes
1
1
1 pair
1
1
1
1
1
1 pack
Task 1
Laboratory stand
Round-bottom three-necked flask, 100 mL
Reflux condenser, connected to water supply
Glass ground joint stopper
Dropping funnel, 50 mL
Oval magnetic stir-bar (big)
Pear-shaped round-bottom flask for distillation,
50 mL
Claisen distillation adapter
Thermometer with fixed ground joint tube
Buchner type fritted glass filter
Rubber spacer for vacuum filtration
Liebig (downward) condenser
Distilling receiver cow
Receiver flask, 10 mL
Receiver flask, 50 mL
Adjustable lab jack lift support
Oval magnetic stir-bar (small)
Plastic beaker, 50 mL, labeled “For the receiver
with the product”
96
2
1
1
6 (with your student code)
1
1
1
2
3
4
5
6
1
7
1
1
1
1
1
1
4 (with your student code)
1
1
1
8
9
10
11
12
13
14
15
16
17
1
IChO Baku: Practical Test
Teflon sleeves for ground tapered joints
Large funnel, 65 mm, with short stem
Joint clips
Grey clamp
Red clamp
Permanent marker
Glass beaker, 25 mL
Plastic container labeled “Used glassware”
Plastic container labeled “Ice bath”
Digital manometer
Cotton wool
Spatula
Glass rod
Ruler
Pencil
12
1
5
1
1
1
1
1
1
1
3
1
1
1
1
18
19
20
Task 2


Laboratory stand
Clamp for burette
Plastic beaker, 100 mL, labeled “Waste”
Glass beaker, 150 mL
Volumetric flask with a stopper, 100 mL
Small funnel, 45 mm
Medium-size funnel, 55 mm
Watch glass
Burette, 25.00 mL, clamped in the stand
Volumetric pipette, 10.00 mL
Graduated pipette, 5.00 mL
Erlenmeyer flask, 150 mL
Graduated cylinder, 100.0 mL
Pasteur pipette
White paper sheet
1
1
1
1
1
1
1
1
1
1
1
2
1
2
1
Task 3
Photometer, 525 nm
Thermostat with adaptor
Spectrophotometer cell with 3.5 cm optical
path length
Magnetic stirrer
Magnetic stir-bar (medium-size)
Netbook with adaptor and mouse
Volumetric flask with a stopper, 100 mL
Graduated pipette, 2 mL
Memory stick 8 Gb
Black magnet
1
1
1
2
2
3
1
1
1
1
2
1
1
4
97
IChO Baku: Practical Test
TASK 1. Tuning bromination selectivity by catalysis
Selectivity of chemical reactions is one of the most challenging problems of contemporary research.
In many cases, reaction conditions and the catalysts applied are keys to achieving high selectivity of
organic reactions. In this task, you will study one of such cases. 3-Methylthiophene can theoretically
be transformed into four monobrominated derivatives T1-T4, which have been actually synthesized
and characterized in detail. Structures of T1-T4 and the values of refractive indexes are given in Table 1.
Table 1. Structures and refractive indexes of monobrominated thiophenes.
Designation
A
B
T3
T4
Br
Structure
S
S
S
nD20
Br
Br
1.5961
S
Br
1.5706
1.5786
1.5795
The selective synthesis of each of T1-T4 can be performed using 3-methylthiophene as the starting
material. T1 and T2 can be obtained by direct bromination using different catalysts, whereas T3 and
T4 are the products of “one pot” multistep synthesis (see Scheme 1).
T2
NBS, CCl4
NBS, CCl4
AIBN cat.
HClO4 cat.
T1
1. 3.5 eq. Br2,
NaOAc,
H2O, 100 °C
S
T4
1. BuLi, TMEDA,
Et2O, rt
2. CBr4, -70 °C
2. Zn dust
T3
O
NBS =
N Br
AIBN = NC
N N
CN
O
TMEDA =
N
N
Scheme 1. Selective synthesis of monobrominated thiophenes.
Q1. Assign the structures given in Scheme 1 with T1, T2 to the structures given in the Table 1.
In this task, you will:
- Synthesize a monobrominated thiophene derivative using one of the catalysts from the list
below;
- Measure the product refractive index (nD)
- Compare the results obtained with literature data and decide on the product structure and
the catalysts given.
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IChO Baku: Practical Test
List of possible catalysts:
HClO4 in CCl4,
AIBN in CCl4
PROCEDURE
- Apparatuses used in this task are shown in Fig. 1 and 2.
- Always equip every joint with the Teflon sleeve. Immediately place every piece of the used
glassware in the corresponding container. Always keep the container tightly closed.
- You can use cotton gloves when handling hot things!
Step 1. Clamp the three-necked flask on the laboratory stand on top of the hot-plate magnetic
stirrer. (Fig.1). Apply the dropping funnel and the reflux condenser to the corresponding necks and
put the big magnetic stir-bar into the flask through the open neck. Ask your lab assistant to switch
on water supply in the reflux condenser (Do not do it yourself!). Transfer NBS quantitatively into the
flask using spatula and big plastic funnel. Transfer ~15 mL of CCl4 into the 25 mL glass beaker. Pour
~2/3 of the solvent volume from the beaker into the flask. Shake the Catalyst and quantitatively add
it through the same plastic funnel into the flask. Add the rest of the solvent from the beaker to the
flask. Close the open neck with the stopper. Put the flask into the ice bath filled with water and ice
to ~ 2/3 of its volume. Start stirring the mixture.
Fig. 1. Set up needed to implement Steps 1-4 of the synthesis. Refer to page 4-5 for the numbers
Step 2. Using the big plastic funnel quantitatively transfer the solution of 3-methylthiophene to
the dropping funnel with tap closed. Apply a piece of the cotton wool to the open end of the dropping funnel and reflux condenser. Under vigorous stirring, add the solution of 3-methylthiophene
dropwise during ~ 3 min. Replace the dropping funnel by a glass stopper. Use the Teflon sleeve. Remove the ice bath. Dry the plate and bottom of the flask with napkin.
Step 3. Wrap up the flask with aluminum foil. Switch on the heater (position 3). Bring up the
mixture to boiling and boil it for 10 min. Prepare the ice bath (~2/3 of the volume) while the mixture
boils.
Step 4. Switch off the heater and carefully (hot!) remove the hot-plate magnetic stirrer aside.
Dip the flask equipped with the condenser and stoppers into the ice bath for 3-5 min. Keep gently
swirling the flask from time to time to provide the faster cooling. Then remove the reflux condenser
and load 0.02 g of K2CO3 using the big funnel applied to the open neck. Close the neck with a glass
stopper and shake the flask several times. Turn off the water supply and unscrew the adaptors of the
tubings from the reflux condenser. Let the residual water leak out of the condenser and immediately
99
IChO Baku: Practical Test
place it into the container for the used glassware. Remove the clamp keeping the flask in the ice
bath.
Step 5. Weigh the 10 mL receiver flask for product with the glass stopper, both marked with
your student code. Write down the value in the answer sheet. Put the small magnetic stir-bar in the
50 mL pear-shaped distillation flask. Screw on the adaptors with tubings to the Liebig condenser and
fix it on the stand with the red clamp. Turn on the water supply yourself and make sure there is no
water leakage.
Step 6. Assemble the distillation apparatus as shown on Fig. 2 supplying all the joints with the
teflon sleeves and clips. First, attach two 10 mL and one 50 mL receiver flasks to the distilling receiver cow. Then connect the vacuum hose to the cow and complete assembling. Fix the apparatus over
the magnetic stirrer to adjust it on height. Use the adjustable lab jack lift support.
Fig. 2. Set up needed to implement Steps 5-10 of the synthesis. Refer to page 4-5 for the numbers
Step 7. Remove the hot-plate magnetic stirrer aside. Insert the fritted glass filter into the Claisen
distillation adapter using the rubber spacer. Turn on the water-jet pump and switch on the digital
manometer. Remove the three-necked flask from the ice bath and dry it with a napkin. Carefully
transfer the reaction mixture from the three-necked flask to the filter (Attention! If you do it too
fast, the mixture can be partially sucked into the curved part of the adaptor). When finished, turn
off the pump and replace the filter with a glass stopper, use the teflon sleeve.
Step 8. Tightly wrap up the flask and distillation adaptor with aluminum foil up to the thermometer joint. Bring back the magnetic stirrer and turn on stirring and heating (position 6). Do not turn
on the water-jet pump! Collect the distilled solvent into the 50 mL receiver. When the solvent distillation is over, turn off heating and stirring and carefully (hot!) remove the hot-plate magnetic stirrer
aside. Replace the receiver flask containing the distilled solvent by a new 10 mL receiver. Close the
50 mL flask with a glass stopper and deliver it to your lab assistant.
Step 9. Remove the foil and put the pear-shaped bottom flask into the ice bath for 2-3 min to
bring the temperature below ambient. Remove the ice bath; dry the flask with a napkin. Bring back
the magnetic stirrer under the flask (Attention! The hot-plate may be still hot!). Turn on stirring.
Wrap up the flask tightly with aluminum foil. Switch on the water-jet pump. When vacuum is stabilized (follow the reading of the digital manometer), turn on the heater (position 6). Observe the ini-
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IChO Baku: Practical Test
tial stage of the targeted product distillation and collect the first 3-5 drops into an attached receiver
flask not labeled with your student code. Then rotate the cow and collect the targeted product into
the receiver flask with your student code. Write down the product boiling point and pressure reading from the digital manometer into the answer sheet.
Step 10. When the targeted product is collected, turn off the heater, remove the foil and carefully (hot!) remove the hot-plate magnetic stirrer aside. Cool down the apparatus to ambient temperature using the ice bath. Ask your lab assistant to disconnect the vacuum line. Disconnect the receiver flask with the targeted product and immediately close it with the glass stopper labeled with your
student code. Do not attempt to drag the teflon sleeve out if it remains in the receiver. Place the
flask into the 50 mL plastic beaker labeled “For the receiver with the product”. Immediately attach a
new receiver instead of the removed one and apply the joint clip. Leave the apparatus as it is.
Step 11. Measure the refraction index (before weighing) following the instruction below. Record
the temperature from the refractometer.
Weigh the receiver with the targeted product closed with the labeled stopper. Calculate the
mass and yield of the product (take the mass of the teflon sleeve equal to 149 mg). The molar masses of 3-metylthiophene and the product equal 98 and 177 g mol-1, respectively.
Q2. Write down all the result in the followingtable .
#
1
2
3
4
5
6
7
Parameter /Characteristics
Mass of the receiver flask with the glass stopper with student code
Mass of the product
Yield of the product
Refraction index for the product
Temperature recorded from the refractometer
Boiling point of the product
Pressure at the Boiling point
Value
Units
g
g
%
°C
°C
mmHg
REFRACTO 30GS – OPERATING INSTRUCTIONS
Fig. 3. Using the Refracto 30GS
101
IChO Baku: Practical Test
1.
2.
3.
4.
5.
6.
7.
To switch Refracto 30GS on, press and hold “ESC” button (1) until the display appears. The
instrument is ready for operation. It switches off automatically if not operated for 10 min.
Clean the cell and glass rod with a napkin wetted with the solvent from the washing bottle
labeled “cleaning solvent”. Dry both with another napkin.
Make sure the sample to be measured has reached ambient temperature and is homogeneous.
Apply 2-3 drops of the sample onto the measuring cell (2) using the glass rod.
To start the measurement press and hold the ok button (3) until the beep.
Take the value of the refraction index and the temperature from digital display (4) and write
down the result in the answer sheet.
Clean up the cell and the glass rod.
Q3. By comparing the obtained and literature data, draw the structure of the product and catalyst given.
Q4. Draw the structure of the 3-methylthiophene-based reactive intermediates behind the selectivity in the case of T1 and T2.
Q5. Write down the product (T1 or T2) formed as a result of direct bromination of 3methylthiophene with NBS under the given conditions / catalyst used.
ZnBr2
Dibenzoyl peroxide
LiBr in AcOH
Visible light or UV light
Q6. In the synthetic pathways to T3 and T4, draw the structures of the compounds formed in the
first steps of each pathways shown on Scheme 1.
TASK 2. Analysis of the solution of a chromium – vanadium alloy
Antiferromagnetic materials show a good prospect in the development of memory devices for ultrahigh-density data storage, the world’s smallest magnetic memory bit using only 12 atoms being one
of prime examples. Vanadium – chromium alloys exhibit antiferromagnetic properties at subzero
temperatures. It is obvious that composition of alloys used in various hi-tech applications should be
accurately controlled.
In this task, you will analyze an aqueous solution simulating the product of digestion of vanadium –
chromium alloy sample. The task consists of two parts:
I. Oxidation of vanadyl (VO2+) to vanadate (VO3-) in the test solution using potassium
permanganate, followed by determination of vanadium (note that chromium (III) is not
oxidized under these conditions).
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IChO Baku: Practical Test
II. Oxidation of the test solution with ammonium persulfate, followed by titrimetric
determination of the total content of vanadium and chromium with Mohr’s salt
(Ammonium iron(II) sulfate).
Procedure
- The amount of vanadium and chromium should be calculated and reported in mg per 100
mL of the test solution.
- Start doing this task with Part A, since you will need time to oxidize the test solution to be
analyzed in Part C.
- The 10.00-mL volumetric pipette has two graduation lines. You should pipette a volume
between the two lines.
Part A. Preparation of the solution for determination of vanadium and chromium total content
1. Transfer a 10.00-mL aliquot of your test solution into the 150-mL beaker and add 20 mL of
1M sulfuric acid using the 25-mL graduated cylinder.
2. Add 6–8 drops of the 0.3% solution of silver nitrate (the catalyst) and heat the mixture on
the hotplate to 70–80°С (position 3), until condensate on the beaker wall appears.
3. Add 20 mL of the 10% ammonium persulfate solution to the heated mixture using the 100mL graduated cylinder.
4. Continue heating and observe the appearance of yellow color, indicating the formation of
dichromate.
Note! You can perform the determination of vanadium (Part B, 1 – 6), while the mixture is being
heated.
5. Keep heating the mixture for 10-15 min (position 3) after appearance of the yellow color to
decompose the excess of ammonium persulfate (the decomposition is over when you see no
small bubbles in the solution).
6. Cool the solution to ambient temperature.
7. Transfer quantitatively the solution from the 150-mL beaker into the 100-mL volumetric
flask, dilute to the mark with distilled water, stopper the flask and mix thoroughly.
Part B. Titrimetric determination of Vanadium
1. Transfer a 5.00-mL aliquot of the test solution into an Erlenmeyer flask using the graduated
pipette.
Note! The 5.00-mL graduated pipette is self-draining.
2. Carefully add 0.03 M potassium permanganate solution dropwise, shaking the flask after
adding each drop until light pink color appears. Make sure that the light pink color is stable.
Remove the excess of potassium permanganate by adding 0.03 M oxalic acid solution dropwise. Shake the flask after each drop until the light pink color changes to pale blue. Let the
solution stand for about 1 min to make sure that the pink color has disappeared completely.
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IChO Baku: Practical Test
3. Transfer 10 mL of the 1M H2SO4 solution into the Erlenmeyer flask using the 25-mL graduated cylinder.
4. Add 2–3 (not more!) drops of the indicator into the Erlenmeyer flask and shake it vigorously.
Let the flask stand for 2–3 min and observe the purple color appearance.
5. Fill the burette with the Mohr’s salt solution. Use the 100-mL plastic beaker labeled “Waste”
to drain the excess of Mohr’s salt solution from the burette, record the initial reading.
6. Titrate the solution in the Erlenmeyer flask with the Mohr’s salt solution until the color
changes to pure light green through brownish-grey one.
7. Take the final reading of the burette. Repeat as necessary.
Q1. Show your results on the answer sheet.
Part C. Titrimetric determination of vanadium and chromium total content in the test solution
1. Wash the 10.00-mL volumetric pipette with distilled water, rinse with the solution prepared
in 100-mL volumetric flask (obtained in part A).
2. Pipette a 10.00-mL aliquot into an Erlenmeyer flask, add 10 mL of 1M H2SO4 solution using
the 25-mL graduated cylinder.
3. Add 3–4 drops of the indicator. Vigorously shake the flask and let it stand for 3–4 min. Observe appearance of red color.
4. Fill the burette with the Mohr’s salt solution. Use the 100-mL plastic beaker labeled
“Waste” to drain the excess of Mohr’s salt solution from the burette, record the initial reading.
5. Titrate the solution in the flask with the Mohr’s salt solution until the color changes to light
yellow-green.
6. Take the final reading of the burette. Repeat as necessary.
Q2. Show your results on the answer sheet.
Part D. Questions and Data Analysis
Q3. Write down the balanced chemical equations for the reactions that take place upon:
a) oxidation of the test solution with potassium permanganate
b) titration of vanadate with Mohr’s salt
Q4. Write down the balanced chemical equations for the reactions that take place upon:
a) oxidation of the test solution with ammonium persulfate
b) titration of the oxidized test solution with Mohr’s salt
Q5. Calculate the a) V(IV) and b) Cr(III) concentrations in the test solution. Calculate the amount of
the metals in mg per 100 mL of test solution.
Q6. This protocol can not be applied to the determination of vanadium and chromium in steels, if the
steel was digested by conc. HCl. Give equations of two reactions to explain the reasons behind.
104
IChO Baku: Practical Test
TASK 3. Kinetic determination of Diclofenac (DCF)
Kinetic methods with spectrophotometric detection for assaying drugs have been intensively developed during the last decade due to a number of obvious advantages, including inherent simplicity,
cost-effectiveness, availability in most quality control laboratories, and improved selectivity. In this
task you will:
 Perform kinetic determination of Diclofenac (DCF) in a medicine by following the progress of
the drug oxidation reaction.
 Determine the reaction order with respect to DCF
Q1. Spectral changes in the course of DCF oxidation with KMnO4 are given in Fig. 4, (1 to 10 reflects
the reaction progress). Complete the table below suggesting which wavelengths can be applied
for photometric kinetic determination of DCF. In each case, indicate the direction of the absorbance changes (denote increasing with  and decreasing with ).
#
1
2
3
4
5
Wavelength, nm
420
480
520
580
610
Fig. 4. DCF oxidation with KMnO4
Yes or No and direction
Procedure
Part A. Assembling of laboratory equipment
Assemble the laboratory equipment as shown in Fig. 5. Connect the photometer (1), 525 nm (fixed
wavelength) and thermostat (2) to the Netbook via USB slots. Connect the thermostat to the cable
labeled “Thermo” to the power supply at your work place via the power adapter. Put the optical
cuvette (3) on top of the magnetic stirrer (4), pass the cuvette through the photometer from aside
(not possible from top down) and place the thermostat over the cuvette from top down (Fig. 5b).
105
IChO Baku: Practical Test
Fig. 5. Laboratory equipment
Hints!
- Plug in your Netbook to the mains before switching on.
- Plug in all the equipment (the photometer and thermostat) before switching on the Netbook. Switch on the mouse.
- If only one window (hereafter referred to as Pattern) instead of two appears after launching the software, quit and re-launch the program.
- Do not unplug ANY device from the USB slot while carrying out the measurements. If it still
happens, you will see a warning on the screen. Quit and re-launch the program.
- If your Netbook falls asleep, click the «Setup» button in the Measurements window on the
absorbance plot pattern when reverting to the measurements.
- In case you see chaotic temperature changes on the screen, stop and re-start the measurement.
Part B. Plotting of the calibration curve
All measurements needed to plot the calibration curve are carried out at 30 °C with constant KMnO4
and H2SO4 initial concentrations. The DCF concentration is varied by using 4 different aliquots (of 0.2,
0.4, 0.6, and 0.8 mL) of the DCF stock solution.
1) Transfer 5 mL of 1M H2SO4 solution using the graduated cylinder and 0.2 mL of DCF stock
solution using the 2 mL pipette into the 100 mL volumetric flask, dilute to the mark with
distilled water, stopper the flask and mix thoroughly.
2) Carry over the flask contents into the cuvette, put the medium-size stir-bar and switch on
the magnetic stirrer. Adjust the stirring speed regulator to the mark shown on Fig. 5a to
provide for intensive mixing.
3) Launch the «Chemistry-Practicum» software on the Netbook. The software will detect the
external devices (sensors) automatically. You will see two plot patterns (that of
absorbance/extinction/optical density, D vs. t, s; and that of temperature, T °C vs. t, s) on the
display.
4) Set the following parameters in the Menu bars of the corresponding plot patterns (Fig. 6):
106
IChO Baku: Practical Test
-
Click the
icon next to the
button («Fixes X-axis maximum on screen») on the absorbance plot pattern. The entire plot will always fit to the screen;
-
Click the
button («Sets the Y range») on the absorbance plot pattern and set the absorbance range (the ordinate axis) from -0.1 to 1.1.
Type “2” (instead of “1”) in the box of the measurements interval on the absorbance plot
pattern.
Choose «Precisely» in the «Precisely/Roughly» window on the temperature plot pattern,
then click on the «T = X» button and set the required temperature of 30 C in the pop-up
window.
Calibrate the photometer by clicking the «Setup» button in the Measurements window on
the absorbance plot pattern.
-
-
Fig. 6. “Chemistry-Practicum” software interface
Note! Setting the parameters (step 4) is needed only prior to the first measurement.
5) Click the
button («Start measure for chosen sensors») to switch on the thermostat and
observe the lamp heating up the solution in the cuvette. Follow the current temperature
reported in the line above the plot. Wait until the thermostat lamp switches off, reflecting
the set up temperature is attained. Stop the measurements by clicking
button (is
activated and turns to red-orange when the measurement is on).
6) Click any part of the absorbance plot pattern to activate it. Take 2 mL of the KMnO4 solution
using the 2 mL pipette. Click the
button («Start measure for chosen sensors») in the
Menu bar of the Measurements window and quickly blow out (press the pipette piston) the
permanganate solution from the pipette into the cuvette.
Note! Make sure the temperature in the cuvette equals 30 C before adding the KMnO4 solution!
107
IChO Baku: Practical Test
7) Observe the progress of the kinetic curve on the screen. Continue measurement for 50 s
after adding the KMnO4 solution, then terminate the measurement by clicking the «Stop
measurements»
button.
8) Save the data by clicking the
button («Export all the data collected in an external file») in
the Menu bar of the absorbance plot pattern, choose the Desktop and type the file name
“DCF2” (change the name to “DCF4”, or “DCF6”, or “DCF8” in the subsequent experiments).
Note!
- Use only the names of the given format!
- Always save the data on your Desktop before starting the next experiment, otherwise the
-
current data set will be lost after the next click on the
button.
Make sure absorbance plot pattern is active when exporting the data. Otherwise, you will
export invalid results. In case no pattern is chosen, you will get a warning.
9) Empty the cuvette into the Waste bottle, wash thoroughly the cuvette with distilled water.
Use black magnet from the outer side of the cuvette to avoid your stir-bar being dropped
into the Waste bottle while washing. Wipe carefully the external surfaces of the cuvette with
the napkin. Also, use the napkin to dab the thermostat lamp.
10) Repeat the steps 1), 2) 5)-9) with the other volumes of the DCF stock solution.
Part C.
1. Studying of the DCF containing medicine (“Control”)
1) Wash the volumetric flask and prepare the mixture as described above using a 0.4 mL
aliquot of the medicine (“Control”) instead of the DCF stock solution.
2) Repeat the steps 1), 2), 5)-9) described in Part B. When saving the data, name the file
“DCFmed”.
3) Repeat the measurement of the “Control” as necessary.
2. Experimental data analysis
1) Open the Excel file on your memory stick in Excel. One by one open your saved data files in
Notepad by double clicking on them on Desktop. Choose Edit/Select All in the Menu bar,
then right click and copy the selected data into the Excel sheet with the corresponding name
(the volume of DCF added or “DCFmed”) and choose Edit/Paste in the Menu bar. You will
see the experimental data on the Excel sheet (time, s, in column A, and absorbance in
column B).
2) Ignore the values before the maximum. Select columns A and B, and plot the data. Use the
“Insert Scatter” icon shown on Fig. 7.
108
IChO Baku: Practical Test
Figure 7. Position of the “Insert Scatter” icon
3) Choose the initial linear section of the remaining curve (15 to 20 data points), apply linear
approximation by adding the linear trend line and bring the parameters to the chart area.
Make sure that the R2 value exceeds 0.98. If needed, decrease the number of the experimental data points plotted removing later data points. Still always search for the most wide
range of the experimental data providing for the target R2 value. Determine the value of the
initial rate of absorbance change, v0.
Note! You will get zero point for this part of the task if less than 12 values are included in the plotted data range.
4) Analyze similarly the experimental data obtained with the other DCF concentrations and
with the medicine solution “Control” (“DCFmed” file).
5) Calculate the DCF concentrations in the reaction mixtures (in mg/L). Write down the DCF
concentrations and initial rates in appropriate cells of the “Results” Excel sheet.
6) Plot the calibration graph on the “Results” sheet and use it to determine the DCF
concentration in the analyzed mixture prepared from the medicine (“Control”). Fill in the
appropriate cells of the “Results” Excel sheet with the coefficients of linear approximation of
the calibration graph. Calculate the DCF concentration in the medicine.
7) Write down the accepted value in the cell F10 of the “Results” sheet.
8) On the “Results” Excel sheet, graphically determine the reaction order with respect to DCF
and write down the exact obtained value in the cell I3.
9) Once finished, save your file and invite your Lab assistant to demonstrate that you have got
experimental data in the Excel file. Sign and get the Lab assistant’s signature.
Note! Only the data saved on the memory stick will be considered as the result of the Task.
109
IChO Baku: Solutions Theoretical Test
Solution to the theoretical problems
Solution to problem 1
1.1. Ammonia
Chlorodifluoromethane
1.2. U
H
S
G
Q = νΔHvap = 21.3 kJ;
Q = νΔHvap = 20.0 kJ;
V
1.3. Ammonia
Chlorodifluoromethane
COP = Q/W = 7.98
COP = Q/W = 4.10
2.1. d
2.2. d
3.
W = νCv(gas)(T2 – T1) = 2.67 kJ
W = νCv (gas)·(T2 – T1) = 4.88 kJ.
2.3. b
CF3CH2F COP = ΔHvap / (Cv(gas)(T2 – T1)) = 2,80
CF3CF=CH2
COP = ΔHvap / (Cv(gas)(T2 – T1)) = 1,59
Yes No
4.1. T0 = 10,4 + 1114 / (4,87 – lg p1) = 264 K
ΔV = (νRT1 / p1) – (xνRT0 / p1) = 6,7 L
Q = ν (1–x)(ΔHvap – RTvap + (Cv(gas)–Cv(liq))(T0–Tvap)) + νCv(gas) (T1–T0) + p1ΔV = 19,8 kJ
or
Q = ν (1–x)(ΔHvap + (Cv(gas)+R–Cv(liq))(T0–Tvap)) + ν(Cv(gas)+R)(T1–T0) = 19,8 kJ
4.2. W = νCv (gas) (T2– T1) = 3,15 kJ
COP = Q/W = 6,3
4.3. The internal energies of the refrigerant are equal in points 0 and 3. Thus,
x·(ΔHvap – RTvap + (Cv (gas) – Cv (liq))(T0– Tvap)) + Cv (liq) (T0– T3) = 0,  T3 = 298 К.
4.4. c
(because the length of 0-1 line decreases) or (because x (see 4.3) increases and less liquid is in the
equilibrium mixture at T0, so less heat Q is necessary to evaporate it!)
Solution to problem 2
1.1. The value of I should increase with the increase of [Ac]0 at [In]0 = const, because the larger fraction of the
intermediate product R will enter the reaction (3b). The maximum value of I will be achieved if all R reacts
in (3b),
 I = 1/β.
1.2. Shilov’s mechanism includes the initial reaction
A + In  R
(3a')
and two competitive reactions
R + Ac  P1
(3b)
R  P2
(3a'')
The rates of conversion of In and Ас are determined by the rates of the reactions (3а') and (3b), respectively:
𝑘(3𝑎′ )[𝐴][𝐼𝑛]
𝑘(3𝑏)[𝑅][𝐴𝑐] 𝑘(3𝑏)[𝐴𝑐]· 𝑘(3𝑏)[𝐴𝑐]+𝑘(3𝑎′′ )
𝑘(3𝑏)[𝐴𝑐]
=
=
=
′
𝑟(3𝑎 ) 𝑟(3𝑎′ )[𝐴][𝐼𝑛]
𝑘(3𝑎′ )[𝐴][𝐼𝑛]
𝑘(3𝑏)[𝐴𝑐]+𝑘(3𝑎′′ )
𝑟(3𝑏)
in steady-state approximation for [R]. We see that the ratio of two rates does not depend on the initial
concentration [In]0 and I will also not depend on it. This gives the straight line parallel to the [In]0 axis on
the graph.
0
[Ac] = const
I = n Ac /n In
I = nAc/nIn
[Ac]0 = const
[In]0
0
[In]
110
IChO Baku: Solutions Theoretical Test
1.3. In this case I will permanently increase with the increase of [Ac]0   at [In]0 = const. The rate of the
reaction (4b) may be so high that conversion of In in reaction (4а) will be negligible.
Hence I  ∞ if [Ac]0   at [In]0 = const.
2.1. The standard Gibbs energy of reaction (5) at 600К is 66 kJ/mol. The equilibrium constant is
-66000/8,314/600
-6
K= e
= 1.8·10 = k5/k-5 .
Reaction is considered at standard pressures of all the reactants and products. The ratio of the rates of
forward and reverse reactions is
r5
r−5
=
k5 [Br][H2 ]
k−5 [HBr][H]
=
k5
k−5
-6
= 1.8·10
2.2. The steady-state condition is the same for both radicals, e.g. for radical Н
d[H]
dt
= k5·[Br][H2] - k-5·[HBr][H] - k6·[H][Br2] = 0
The concentrations of all the neutral molecules are the same (they correspond to the pressure of 1 bar),
[H]

[Br]
=
k5 [H2 ]
k−5 + k6
=
k5 /𝑘−5
1 + k6 /𝑘−5
=
1.8·10−6
1 + 10
-7
= 1.6·10
The Gibbs energy of reaction (5) under such conditions is:
G(5) = G°(5) + RT·ln
[H][Hbr]
-3
-7
= [66 + 8.314·10 ·600·ln(1.6·10 ] kJ/mol= -12 kJ/mol
[Br][H2 ]
The ratio of rates is:
r5
r−5
2.3. b
=
k5 [Br][H2 ]
k5
=
k−5 [HBr][H] k−5
·
[𝐵𝑟]
[𝐻]
=
k5
k−5
·
1 + k6 /𝑘−5
[k5 /𝑘−5 ]
=1+
k6
k−5
= 11,3
(Arguments – not required from the student. The statement (β) is true beyond any doubt
rbeob = r5 – r–5 = r5 · (1 -
r−5
r5
) = r5 · (1 - eG/(RT))
ΔG is positive in (2.1) and negative in (2.2), the expression in brackets consecutively is negative in
(2.1) and positive in (2.2). The observed rate robs changes its sign from negative to positive due to
chemical coupling as r5 is always positive.
The statement (a) may not be true as r5 = k5[H2][Br] and change of [Br] is not specified in the problem. k5 and [H2] are the same in (2.1) and (2.2). The forward reaction in (2.2) may be faster or slower.
The statement (c) may not be true for the similar reason.)
3.1. According to the Second law the following condition has to be met:
-7
-1
-1
GSyst/T = G(7) · r7 + G(8) · r8 ≤ 0  r7 ≤ r8 · (-G(8))/ G(7) = (481.2/46) · 6.3·10 mol L min
This is the maximum possible rate of the coupled reaction.
3.2. In this case Gibbs energies for both reactions will change:
-1
-1
G'(7) = G(7) – RT·ln2 = (46.0 – 1.8) kJmol = 44.2 kJmol
-1
-1
G'(8) = G(8) + RT·ln2 = (-481.2 + 1.8) kJmol = -479.4 kJmol
The observed rate of the reaction (8) will be decreased due to the increase of the rate of reverse reaction
(8). However with such a huge negative Gibbs energy of the reaction (8) the rate of reversed reaction is
negligible. The maximum rate of the reaction (7) is
-8
-1
-1
-7
-1
-1
r7 ≤ r8 · (-G(8'))/ G(7') = (479.4/44.2) · 6.0·10 mol L min = 6.5·10 mol L min
The relative increase of maximum rate is practically negligible:
-7
-7
r7'/r7 = (6.5·10 )/ (6.3·10 ) = 1.03
 r7(max. 3.2) / r7(max. 3.1) = 1.03
111
IChO Baku: Solutions Theoretical Test
Solution to problem 3
1.1. Cooperation: β > 1
1.2. K =
[H𝐺2 ]
[HGb ][G]
=
Competition: 0 < β < 1
[H𝐺2 ]
[HGa ][G]
·
[H𝐺𝑎 ]
[HGb ]
= β·Kb ·
𝐾𝑎
Kb
Independence: β = 1
= β·Ka
2.1. Kb = 2 Ka  [HGb] = 2 [HGa]
Material balance with respect to H: [H] + [HGa] + [HGb] +[HG2] = [H]0 = 1 mol/L
or
0.1 + 3[HGa] + [HG2] = 1 mol/L
Material balance with respect to G: [G] + [HGa] + [HGb] +2[HG2] = [G]0 = 2 mol/L
or
0.5 + 3[HGa] +2 [HG2] = 2 mol/L
Solving the system of two equations, we find: [HGa] = 0.1 mol/L, [HG2] = 0.6 mol/L,
hence [HGb] = 0.2 mol/L)
Ka =
[H𝐺𝑎 ]
[H][G]
=
0.1
0.1 ·0.5
=2
β=
[H𝐺2 ]
[HGa ][G]𝐾𝑏
=
0.6
0.1 ·0.5·4
=3
2.2.
H
HGa
HGb
HG2
2.3. 1)
[HG2] = [HGa] + [HGb] = 3[HGa]
[HG2 ]
[HGa ][G]
2)
3)
4)
2.4. a)
= β∙Kb = 12
112
= 12
[G] = 0.25 mol/L
Material balance with respect to H: [H] + 3[HGa] + [HG2] = 1 mol/L
[H] + 6[HGa] = 1 mol/L
[H] + 12[H][G] = 1 mol/L
[H] = 0.25 mol/L.
[HGa] = Ka[H] [G] = 0.125 mol/L.
[HG2] = 3[HGa] = 0.375 mol/L.
Material balance with respect to G: [G]0 = [G] + 3[HGa] + 2[HG2] = 1.375 mol/L
n0(G) = 1.375 mol/L
β = 0:
In this case, no HG2 is formed.
Material balance with respect to H: [H] + [HGa] + [HGb] = 1 mol/L
Material balance with respect to G: [G]0 = [G] + [HGa] + [HGb] = 2 mol/L
Equilibrium constant: Ka =
b)
3
[G]
[HGa ]
[H][G]
[H] + 3[HGa] = 1 mol/L
[G] + 3[HGa] = 2 mol/L
=2
Solving the system of three equations, we get:
[H] = 0.129 mol/L
[G] = 1.129 mol/L [HGa] = 0.290 mol/L
[HGb] = 0.580 mol/L
[HG2]
=0

In this case, formation of HG2 is practically irreversible, so only HG2 is present in the
solution.
[H] = 0
[G] = 0
[HGa] = 0
[HGb] = 0 [HG2] = 1 mol/L
IChO Baku: Solutions Theoretical Test
Solution to problem 4
1.
2.
The precipitate X2 formed by addition of barium chloride in acidic medium is barium sulfate BaSO 4.
The precipitate X3 formed by addition of silver sulfate is silver chloride AgCl.
The yellow precipitate X4 formed by addition of alkali can be mercury oxide HgO or silver phosphate
Ag3PO4. The mole ratio n(X4) : n(X2) is 0.931 for n(HgO) : n(BaSO4) which is not valid and 1.798 for
n(Ag3PO4) : n(BaSO4) which gives 2.4 being multiplied by 4/3. So, the molar ratio is nAg3PO4) : n(BaSO4) =4
:3 which corresponds to n(P) : n(S) = 4:3, i.e. to formula of X1 P4S3.
X1 = P4S3
X2 = BaSO4
X3 = AgCl
X4 = Ag3PO4
The gas evolved has a molar mass 1,586 · 29 g/mol = 46 g/mol that is NO2
P4S3 + 38 HNO3  4 H3PO4 + 3 H2SO4 + 38 NO2+ 10 H2O
H2SO4 + BaCl2  BaSO4+ 2 HCl
Ag2SO4 + 2 HCl  2 AgCl + H2SO4
BaCl2 + Ag2SO4  BaSO4 + 2 AgCl
H2SO4 + 2 NaOH
 Na2SO4 + 2 H2O
2 H3PO4 + 6 NaOH + 3 Ag2SO4  2 Ag3PO4 + 3 Na2SO4 + 3 H2O
3.
Phosphorus sulfide P4S3 is a molecular cage
4.
a) P4S3 + 8 O2
 2 P2O5 + 3 SO2
b) P4S3 + 16 H2SO4  4 H3PO4 + 19 SO2 + 10 H2O
c) 3 P4S3+ 16 KClO3 16 KCl + 6 P2O5 + 9 SO2
Solution to problem 5
1.1. C6H12O6 + 2 Cu + 5 OH  C6H11O7 + Cu2O+ 3 H2O
2+
–
–
1.2. 2 CuSO4 + 4 KI  2 CuI + I2 + 2 K2SO4
KI +I2  KI3
2 Na2S2O3 + I2  2 NaI + Na2S4O6
or
or
or
2 Cu + 4 I  2 CuI + I2
–
–
I + I2  I3
2–
–
2–
2
S2O3 + I2  2 I + S4O6
2+
–
1.3. Total amount of copper(II):
50.00 mL · 0.0400 mol/L = 2.0000 mmol
Obviously, there is an excess of iodine and the remaining iodine was titrated with sodium thiosulphate:
11.87 mL · 0.05078 mol/L = 0.6028 mmol
2.0000 mol - 0.6028 mol = 1.3972 mmol of copper(II) was required to oxidize the sugars.
2+
n(cugars) = ½ · n(Cu ) = 0.6986 mmol
c(sugars) = 0.6986 mmol/10.00 L = 0.06986 mol/L
mass content = 180.16 g/mol · 0.06986 mol/L = 12.6 g/L
1.4 KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O
–
–
+
IO3 + 5I + 6H = 3I2 +3H2O
–
–
–
Only glucose was oxidized with iodine: C6H12O6 + I2 + 3OH  C6H11O7 + 2I + 2H2O
or
113
IChO Baku: Solutions Theoretical Test
Na
3NaOH
I2
2NaI
2H2O
2 Na2S2O3 + I2  2 NaI + Na2S4O6
1.5. b
–
1.6. n(I2) = 3n(IO3 ) = 3·0.01502 M · 10 mL = 0.4506 mmol
n(I2) = n(glucose) = 0.4506 mmol
c(glucose) = 0.4506 mmol/10.00 mL = 0.04506 mol/L
Mass content of glucose = 180.16 g/mol · 0.04506 mol/L = 8.12 g/L
Mass content of fructose = (12.6 – 8.12) g/L = 4.48 g/L
1.7. 0.2 L·8.12 g/L = 1.6 g of digestible carbohydrates, that is 0.14 BEU.
2.1. a)
b)
c)
d)
2.2.
O + p  n + …
18
… + 21𝐷  F + 
19
20
F + 21𝐷  F + …
16
18
O + …  F + p + n
18
18
F
Ne
p

20
11
Nucleus
Decay mode
20
С
+
β
17
F
β
14
F
+
β
C
–
β
2.3. The initial ratio of radioactivities:
A0 ( 18F)
A0 ( 17F)
=
η( 18O  18F )
η(( 16O  17F )
·
χ(H2 18O)
χ(H2 16O)
= 144.7 ·
χ(H2 18O)
χ(H2 16O)
After 5 minutes the ratio changed due to radioactive decay of fluorine:
A300 ( 18F)
A300
( 17F)
χ(H2 18O)
χ(H2 16O)
=
𝑙𝑛2
)
109.7
𝑙𝑛2
A0 ( 17F)·exp(−300· )
65
A0 ( 18F)·exp(−5·
= 23.75 ·
A0 ( 18F)
A0
( 17F)
= 3437 ·
χ(H2 18O)
χ(H2 16O)
= 10
5
= 29
18
Mass fraction of H2 O ist
29 ·20
18
ω(H2 O) =
29 ·20+18
= 0.97 ≙ 97 %
2.4. During the synthesis. the radioactivity will decrease:
A3.5 = A0 · exp(- 3.5) = 581.3 MBq
η = 528.2/581.3 = 0.909 ≙ 90.9 %
2.5. Radioactivity is excreted by radioactive decay and through the excretory organs (e.g. kidneys). The excretion process may be considered as two competitive first-order reactions. Activity after one hour is:
A60 = A0 · exp(-(1 + 2)·t) = 450 · (-(exp(
114
ln2
109.7
+
ln2
) · 60) = 218 Bq
120
IChO Baku: Solutions Theoretical Test
Solution to problem 6
1.
A
B
C
E
X
Y
D
2.
F
G
I
J
3.
Solution
1-2. From the given contents of C, H, and N we can calculate the molecular formulas of A, B and E as (C5H9N)n.
This formula corresponds to isomeric piperideines (piperidines containing one double bond). However, piperideines themselves have no 3-fold axis of symmetry. Thus, we conclude that A is a symmetric trimer of piperideine. B is another trimer of piperideine (B is diastereomer of A). Analysis of both Scheme 1 and Scheme 2
enables us to conclude that X and Y are piperidine derivatives. Therefore, compound E should be also piperidine derivative; most probably, this is a monomeric piperideine. It is consistent with the fact that E is an intermediate in both the formation of A, B and their transformation into compound C. The easy trimerization of E
and its reactivity against HCN demonstrates the presence of a highly polarized bond. From thee isomeric piper1
ideines only  -piperideine (3,4,5,6-tetrahydropyridine) has a highly polarized C=N bond; both the spontaneous trimerization and reaction with HCN seem to be impossible for other isomers. Indeed, the chlorination of
piperidine produces N-chloropiperidine which eliminates HCl under the treatment with base affording 3,4,5,6tetrahydropyridine.
The compound A is characterized by 3-fold axis of symmetry. It is possible only if A is an all-cis trimer of E.
Compound B is diastereomer of A, i.e. B is a cis,trans-trimer of E.
115
IChO Baku: Solutions Theoretical Test
The compound C is the product of addition of HCN to a C=N bond of  -piperideine E. Accounting for polarization of this bond, it is 2-cyanopiperidine. It was treated with t-BuOCl and base analogously to the treatment of
piperidine at the first step of the discussed synthesis. It is reasonable to conclude that the same reagents
should lead to the same type of transformation, i.e. D is 2-cyano-3,4,5,6-tetrahydropyridine. This conclusion is
consistent with N for compound D. If D has two nitrogen atoms, the molecular weight of D is 108. It corresponds to the molecular formula C6H8N2. On the contrary to E, this molecule is stable. This stabilization is provided by the conjugation between the C=N and CN bonds; this fact enables to discriminate 2-cyano-3,4,5,6tetrahydropyridine from unstable 2-cyano-1,2,3,4-tetrahydropyridine where such stabilization is absent.
1
Methylmagnesium iodide can react with carbon atoms of both C=N and CN bonds. The triple CN bond is
more polar; a positive charge on the carbon atom of the nitrile group is higher. Attack on this carbon atom has
also lower steric demands. Therefore, MeMgI should react with CN bond faster than with C=N bond. Moreover, this conclusion can be made without this analysis as attack onto C=N bond should produce 2-methyl-2cyanopiperidine. The tautomeric equilibrium for this compounds is impossible.
The hydrolysis of the formed imine anion affords a ketone moiety. According to the problem, this molecule
occurs in equilibrium with its tautomer; Y being the kinetic product and X being the thermodynamic product.
The keto-enol tautomerism is inappropriate here as enols are usually much less stable than ketones. However,
the imine-enamine tautomerism is also possible in this molecule. Both tautomers have the conjugation between two double bonds; the enamine tautomer seems to be more stable due to the hydrogen bond between
1
N–H group and the proximal ketone function. Moreover, compound D is a  -piperideine derivative. The kinet1
ic product of its transformation should be  -piperideine derivative too. It is compound Y. Its enamine tautomer is compound X. The analysis of Scheme 2 supports this conclusion.
Similarly, tautomer of the compound D is 2-cyano-1,4,5,6-tetrahydropyridine (G). To form the mixture of compound D and its tautomer G, intermediate F should have the possibility to react with water by two ways (by
carbon atom and by nitrogen atom). It is possible if F contains an anion obtained by deprotonation of D. The
coloring of the reaction mixture is an additional argument in favor of the conjugate anion formation.
3. It is not necessary for the students to know the first reaction. The comparison of molecular formulae of
compound I and the starting compound H demonstrates clearly that this transformation leads to the substitution of an oxygen atom by a CH2 group. There are two possibilities for this substitution: 1) the transformation
of the ester group into enol ether and 2) the transformation of the less reactive carbamate function into the
116
IChO Baku: Solutions Theoretical Test
corresponding 1-alkoxy-1-aminoalkene. In the first case, the hydrolysis of I should produce 2-acetylpiperidine
(or its N-Boc derivative). In the second case, the product of hydrolysis should be 1-acetylpipecolinic acid (or its
ethyl ester). From the Scheme 1 it is clear that X and Y have no acyl substituent at the nitrogen atom. Moreover, the method of synthesis of X and Y described in Scheme 1 does not correspond to the presence of an ester
function. Oxidation cannot remove Boc function. So, compound J is 2-acetylpiperidine. Its oxidation produces
2-acetylpiperideines X and Y.
Solution to problem 7
1.
XW
XS
2.
B
C
D
E
F
G
3. The number of possible stereoisomers of A: 4
4.
I
J
K
L
Solution
2. The first step is acylation of nortropanol. The comparison of molecular formula of C with the formula of A
shows that transformation of B into C results in a loss of two hydrogen atoms. It is possible to conclude that
this step is alcohol oxidation to ketone (this oxidation was discussed in Preparatory problem). Therefore, the
first step is acylation of nitrogen atom. During the transformation of E into F N,O-dimethylhydroxylamine substitutes –OH group activated by transformation into chloride after treatment with SOCl 2. The comparison of
molecular formulae of F and pelletierine shows that the formation of pelletierine ((S)-1-(piperidin-2-yl)propan2-one) includes the loss of Cbz protecting group (its substitution by hydrogen atom) and substitution
117
IChO Baku: Solutions Theoretical Test
of -N(CH3)OCH3 fragment by methyl group. Analysis of all these steps enables to conclude that G is Cbzprotected pelletierine, E is 2-(N-Cbz-piperidin-2-yl)acetic acid and F is its N-methoxy-N-methylamide. Acid E is
formed from the compound D by addition of two hydrogen atoms (compare molecular formulae of D and E). In
turn, compound D is formed by introduction of one oxygen atom into ketone C. Even without knowledge of
these reactions it is possible to conclude that oxidation of C produces lactone (intramolecular ester) corresponding to acid E (the Baeyer-Villiger oxidation). Treatment of this lactone with Et 3SiH and BF3OEt2 leads to
the cleavage of the ester and reduction of semiaminal moiety –NCH(R)OH to the corresponding saturated
amine. Another way for solution is a comparison of structure of C with structure of pelletierine. This comparison shows that C(O)–CH bond should be broken. Therefore, during C-to-D transformation oxygen should be
introduced into this bond. Product contains the C(O)–CH3 and the CH–H fragments. Therefore, from two possible ways of adding two hydrogen atoms we should select the way that leads to the CH–H fragment and –COOH
moiety. Other arguments are similar to the above ones.
3. Nortropanol A has 3 chiral centers. However, due to bicyclic structure of A only 4 stereoisomers can exist.
4. The knowledge of molecular formula of pelletierine allows to conclude that the last step is the removal of
Boc protecting group. In other words, compound L is Boc-protected pelletierine. The step of K formation is,
evidently, acylation of nitrogen atom in 2-allylpiperidine. So, K-to-L transformation is the oxidation of terminal
alkenyl group into the corresponding ketone (the Wacker process). The formula of the intermediate, the structure of which is given in the scheme, enables to conclude that the base-induced I-to-J transformation is the
elimination of HBr and the next step is the removal of tert-butylsulfinyl group. Then, the J formation includes a
cyclization producing piperidine ring. Therefore, we can write the following structures of compounds I–L.
118
IChO Baku: Solutions Theoretical Test
Solution to problem 8
1a.
n(C) > n(O) 
n(C) < n(O) 
n(C) = n(O) 
Data insufficient 
1b. With regard to 1a, three variants (n(H)>n(C)>n(O), n(C)>n(H)>n(O), and n(C)>n(O)>n(H)) are possible for
X1. For each inequality, one can write down the corresponding formula using elements of a geometric
progression (q is the progression common ratio), equations for calculation of mass fractions of carbon
and its roots
Inequality
Formula
Equation
n(H)>n(C)>n(O)
СqnHq2nOn
n(C)>n(H)>n(O)
Сq2nHqnOn
n(C)>n(O)>n(H)
Сq2nHnOqn
12.01·𝑞·𝑛
12.01·𝑞·𝑛 +1.008·𝑞2 ·𝑛 + 16.00·𝑞·𝑛
12.01·𝑞 2 ·𝑛
12.01·𝑞 2 ·𝑛 +1.008·𝑞·𝑛 + 16.00·𝑞·𝑛
12.01·𝑞 2 ·𝑛
12.01·𝑞 2 ·𝑛 +1.008·𝑛 + 16.00·𝑞·𝑛
1. root (q1)
2. root (q2)
= 0.5453
2.00
7.93
= 0.5453
–1.21
1.32
–0.06
1.66
= 0.5453
There is only one positive integer root, thus the empirical formula is X1 = C2H4O
1c. Since (1) and (2) are the reaction equations, one can write down the formula of X as C2nH4nOn + 2H  1O =
C2nH4n+2On-1. With an account for the known mass fraction of hydrogen:
1,008·(4𝑛+2)
12,01·2𝑛 +1,008·(4𝑛+2)𝑛 + 16,00·(𝑛−1)
= 0.1194
 n = 3 and X = C6H14O2
1d. X2 is formed from X composed of three elements (C, H, and O) via a monooxygenase catalyzed reaction:
n(C) : n(H) : n(O) =
100−51.56−9.74
12
:
9.74
:
51.56
1.008 16.00
= 1 : 3: 1
Since the number of hydrogen atom is necessarily even, the molecular formula of X2 is C2H6O2. Other variants
with a higher even number of hydrogen are not valid. Ethylene glycol (structure: HO-CH2-CH2-OH) is the only
stable substance with the molecular formula deciphered above.
1e.
2.
Number of graph: 1
(The number of kinks of the curve depends on that of urinations per day (4 to 6 times in norm); the amplitude of 2-butoxyacetic acid content in organism can vary; the metabolite is always (even after urine
passage) found in organism in nonzero concentration.)
3.
Dioxygenase incorporates two oxygen atoms in vicinal positions of the substrate, which can be followed
by chemical bonds reorganization. The empirical formula of the hydrocarbon Р is C7H8
n(C) : n(H) = (91.25/12.01) : ((100.91.25)/1.008) = 7 : 8  P = C7H8 . Toluen
The molar mass of P3 equivalent containing acidic group(s) is 100 : (6.41 · 0.100) g/mol = 156 g/mol.
Two dioxygenase steps suggest the composition of C7H8O4.
P3 must be a monocarboxylic acid if it still contains seven carbon atoms. Fragments containing a СH 3CO–
group (or a СH3CH(OH)– group further transforming into СH3CO– one) are involved into the iodoform reaction. This suggests splitting of the benzene moiety during the second oxygenase step at the carbon
connected to the methyl group.
119
IChO Baku: Solutions Theoretical Test
P
4.
120
P1
P2
P3
Thorough consideration of the transformations brings to the conclusion that Y1, Y2, and Y3 are anions of
the same charge. Dioxygenase reaction suggests oxygen as Y4. The identity of the first two steps shows
that Y1 contains by one oxygen atom more than Y2 (Т1 is phenol). So, Y2 must have a composition of
nY3O2 . Formally Y3 is a halogenide or chalcogenide. Still silver chalcogenides, bromides and iodides do
not dissolve even in concentrated ammonia solutions. Fluorine does not afford stable oxygen-containing
anions. Thus all Y particles contain chlorine.
There are four ways of pyrocatechol ring splitting as a result of the dioxygenase reaction, still only one of
these leads to the product without identical functional groups:
Y1
Y2
Y3
–
Cl
Y4
T1
T2
About the History of the IChO
About the history of the International Chemistry-Olympiads
The idea of chemistry olympiads was born 1968 during an Czechoslovakian national olympiad that was attended by observers from Poland and Hungary. These three countries participated in the first IChO 1968 in Prague. The number of teams attending the IChO in the following years is shown in the plot below.
Number of teams attending the IChO
80
of Teams
Number
of Teams
Number
70
60
50
40
30
20
10
0
1970
1980
1990
2000
2010
Year
Yearof0fOlympiad
Olympiad
The participating countries are shown in the following table.
121
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
Argentina
Armenia
Australia
Austria
Azerbaijan
Belarus
Belgium
Brazil
Bulgaria
Canada
China
Chinese Taipei
Costa Rica
Croatia
Cuba
Cyprus
Czech Rep.
Czechoslovakia
Denmark
DDR
Egypt
El Salvador
Estonia
Finland
France
fYROM (Macedonia)
Georgia
Germany
Greece
Hungary
Iceland
India
Indonesia
Iran
Ireland
Israel
 Country
Year 
122
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
+ + + + +
o + + + + + + + + + +
+ + + + + + + + + + + + + + + + +
o
+ +
+ + + + +
+ + +
+ + +
+
• +
+ +
o
+ +
+ + + + + + + + + + + + + + + +
+ + + + + + + • + + + + + + + + + + + + +
o o + + + + + + + + +
+ + + + + + + +
+ + +
+ o + + + + +
o + +
• + + + + + + + • + + + + + + + • + + + + + +
+ + + + + + + + + +
o + + + + + • + + + + + + + + + + + • +
+
+
•
+
o o +
+ + + +
+ • + +
+ + + +
+ + + +
o o
+ + + + + +
+ + + + + + + +
+ + + + + + +
+
+ + + + + + + +
+ + + + + +
o + + + + + + + + + + • + + + + + + + + + + +
o + + + + + + + + + • + + + + + + + + +
o + + + + + + + + + • + + + + + + + + + + + + + + +
+ + + + + + + + + + + + + + +
+ + • + + + • + + + + + + + + + + + • + + + + + + + + + + + +
o o +
o + + +
+ + + + + + +
o o + +
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
Argentina
Armenia
Australia
Austria
Azerbaijan
Belarus
Belgium
Brazil
Bulgaria
Canada
China
Chinese Taipei
Costa Rica
Croatia
Cuba
Cyprus
Czech Rep.
Czechoslovakia
Denmark
DDR
Egypt
El Salvador
Estonia
Finland
France
fYROM (Macedonia)
Georgia
Germany
Greece
Hungary
Iceland
India
Indonesia
Iran
Ireland
Israel
 Country
Year 
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
+ + + + +
o
+ + + + +
+ + + + +
+ + + + +
+ + + + +
+
o
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
•
+
+ + + + + ++ + + + + + + + + +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
++ + +
++ + +
++ + +
+ + +
• + + +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
+ +
+ +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
o
+
+
+
+
o
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ + +
+ + +
+
+ + +
+ + +
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
• + + + + ++ + + + + + + + + +
o o + + + ++
+
o
+ + + + + ++ + + + + +
+ + + + + ++ + + + + +
+ + + + + ++ + + + + +
o
+ •
• +
+ +
+ +
• + +
+ + +
+ + +
+ + +
o
0 0 0 0 0
0 1 2 3 4
+
+
+
o
+
+
+
+
+
+
+
o
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
o
0
5
+
+
+
+
+
+
+
+
+
0
6
+
+
+
+
+
+
+
+
+
0
7
+ + +
+ + +
• + +
+ + +
+ + +
+ + +
+ + +
+ + +
+ + +
0 0 1
8 9 0
+
+
+
+
+
+
+
+
+
1
1
+
o
+
+
+
o
o
+
+
+
+
+
+
+
+
+
1
2
+
+
+
+
+
o
+
+
+
+
+
+
+
+
+
1
3
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1
4
o
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
1
5
123
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
Italy
Japan
Yugoslavia
Kazakhstan
Kenya
Korea
Kuwait
Kyrgyzstan
Liechtenstein
Latvia
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
+ + + + + o o + + + + + + + + + + + + +
+ + + +
+ + + + + +
o o
+ + + + +
o o +
+ + + + + + + + +
+ + + + + + + + +
Lithuania
Malaysia
Mexico
Moldova
Mongolia
Montenegro
Netherlands
New Zealand
Nigeria
Norway
Pakistan
Oman
Peru
Philippines
Poland
Portugal
Romania
GUS/Russ.Fed
Saudi Arabia
Serbia
Singapore
Slovakia
Slovenia
South Africa
Spain
 Country
Year 
124
o o + +
o
+ + + + + + + +
+ + + + + +
+ + + + + + + +
+ + + + + + • + + + + + + + + + + + + +
+ + + + + + + +
o + + + + + + + + + + + + • + + + + +
o
+ • + + + + + + + • + + + + + + + + + + + + • + + + + + + + +
+ + + • + + + + + + + + • + + + + + + + + + + + + + + + +
+ + + + • + + +
o +
o
+ + + + + + + + + +
+ + + + + + +
+ + + + + + + + +
+ + + +
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Italy
Japan
Yugoslavia
Kazakhstan
Kenya
Korea
Kuwait
Kyrgyzstan
Latvia
Liechtenstein
Lithuania
Malaysia
Mexico
Moldova
Mongolia
Montenegro
Netherlands
New Zealand
Nigeria
Norway
0man
Pakistan
Peru
Philippines
Poland
Portugal
Romania
GUS/Russ.Fed
Saudi Arabia
Serbia
Singapore
Slovakia
Slovenia
South Africa
Spain
++ +
o
o
++ +
o
++ +
++ +
 Country
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
Year 
+ + ++ + + + + + + + + +
+ + ++ + + + • + + + + +
+ + ++ + + + + + + + + +
+ + +• + + + + + + + + +
+ + ++ + + + + + + +
++ + + + ++ + + + + + + + + +
++ + + + ++ + + +
o
++ + + + ++ + + +
o ++ + +
++ + + + ++ + + +
o
++ • +
++ + +
++ + +
o o
++
o
++
++
+
o
+
+
+
+
+
+
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o
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+
o
+
+
+
o o + + + + + + +
o o + + + + + + + +
o o
+ ++ + + + + + + +
+ ++ + + + + + + +
o
o o + +
+ ++ + + + + + + +
o
o o + + + + + + + +
+ + + + + + + + +
+
+
+
+
+
+
+
o
+
+
+
+
+
+
+
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+
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o
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o
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o
+
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+
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+
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+
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o
+
+
+
+ + + +
+ + + +
+ + + +
• + + +
+ o o
o
++ + + + ++ + + + +
++ + + + ++ + + + +
++ + + + ++ + + + +
+
o
+
+
+
+
+
+
+
+
o
+
+
+
+
o
+
+
+
+
+
+
+
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+
+
+
+ +
+ +
+ +
• +
+ +
+ +
+ +
+ +
+ +
o
++ + + + ++ + + + + + + + +
125
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
Sweden
Switzerland
Syria
Tajikistan
Thailand
Turkey
Turkmenistan
UdSSR
Ukraine
United Kingdom
United States
Uruguay
Uzbekistan
Venezuela
Vietnam

Country
Year 
126
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9
+ + + + + + + + • + + + + + + + + + + + + + + + + +
o + + + + + + + + + + + + +
o + + + + + + + + + +
o +
o + + + + + +
o
+ • + + + + + + • + + + + + + + + + +
+ + + + + +
o o + + + + + + + + + + + + + + + + +
o o + + + + + + + + • + + + + + + +
o o +
o
o
+ + + +
+
6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9
8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6
+
+
9
7
+
+
9
8
+
+
9
9
About the History of the IChO
Participating Delegations
•= host. + = participant. o = observer
Year 
Country 
Sweden
Switzerland
Syria
Tajikistan
Thailand
Turkey
Turkmenistan
UdSSR
Ukraine
United Kingdom
United States
Uruguay
Uzbekistan
Venezuela
Vietnam
 Country
Year 
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
+ + + + + ++ + +
+ + + + + ++ + +
o
o o + ++ + +
+ + + + + ++ + +
+ + + + + ++ + +
o o + + + + + +
+ + + +
+ + + +
o + + +
+ + +
+ + + +
+ + • +
+ + + +
+ + + + + ++ + + + + + +
+ + + + + ++ + + • + + +
+ + + + + ++ + + + + + •
+ + + + + ++ + + + + + +
+
+
0
0
+
+
0
1
+
+
0
2
+
+
0
3
+
+
0
4
+
+
0
5
+
+
0
6
+
+
0
7
+
+
0
8
+
+
0
9
+
+
1
0
o
+
+
1
1
o
+
+
1
2
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+ + +
+ + +
+ + +
+ + +
+
+
+
1
3
+ +
+
• +
1 1
4 5
127
About the History of the IChO
Inofficial ranking since 1974
(set up by adding the points of the teams. up to position 50)
1974 1975 1976 1977 1978
IChO held in RO H DDR CS PL
1
SU SU DDR CS SU
.
RO
H
SU SU PL
.
CS PL
H
H
D
.
H
BG PL PL DDR
5
PL RO
A
S
CS
.
DDR DDR RO
A
H
.
BG
S
BG
D
A
.
YU CS CS DDR RO
.
S
A
S
RO
S
10
D*
D
D
BG BG
.
YU YU YU TR
.
B
B
B FIN
.
.
15
* hors concourse
.
.
.
.
20
.
.
.
.
25
.
(List of abbreviations see page 132)
128
1979
SU
PL
SU
RO
CS
A
S
H
D
BG
FIN
DDR
1980
A
PL
D
DDR
H
A
RO
BG
CS
S
FIN
NL
I
B
1981
BG
H
CS
PL
BG
A
D
DDR
RO
SU
NL
FIN
S
F
I
1982
S
CS
D
PL
NL
A
SU
H
BG
DDR
S
F
FIN
N
RO
DK
YU
I
1983
RO
RO
SU
D
CS
H
A
F
DDR
PL
NL
BG
GB
N
DK
FIN
S
I
YU
1984
D
D
CS
SU
H
A
GB
PL
USA
RO
DK
S
NL
FIN
F
BG
N
I
GR
YU
B
1985
CS
SU
CS
D
A
NL
H
DDR
PL
USA
F
GB
RO
BG
N
S
FIN
YU
B
GR
DK
C
1986
NL
NL
PL
D
SU
A
USA
H
BG
F
RO
CS
GB
S
DDR
CDN
N
DK
B
FIN
GR
1987
H
SU
RC
RO
CS
D
F
GB
PL
H
DDR
NL
USA
BG
A
S
FIN
N
DK
I
GR
KWT C
YU
B
YU
CDN
CH
1988
FIN
RC
D
USA
PL
GB
DDR
N
RO
H
SU
I
NL
BG
CS
AUS
SGP
F
A
FIN
CDN
DK
C
S
B
CH
KWT KWT
About the history of the IChO
1989
1990
IChO held in DDR
F
1
DDR
RC
.
D
PL
.
RC
D
.
BG
USA
5
SU
CS
.
H
RO
.
PL
F
.
RO
A
.
CS
DDR
10
I
H
.
NL
GB
.
GB
I
.
A
AUS
.
USA SGP
15
S
NL
.
F
N
.
N
DK
.
AUS
T
.
CDN FIN
20
DK CDN
.
FIN
BG
.
B
C
.
C
S
.
GR
CH
25
CH
B
.
KWT GR
.
KWT
.
CY
.
30
.
.
.
.
35
.
.
.
.
40
.
.
.
.
45
.
.
.
.
50
1991
PL
RC
RO
H
PL
NL
USA
I
D
N
GB
CS
SU
A
AUS
DK
SGP
CDN
BG
F
S
T
CH
LV
LT
FIN
C
GR
B
CY
SLO
1992
USA
RC
H
PL
USA
A
GUS
D
RO
F
I
SGP
CS
AUS
NL
DK
ROK
GB
CH
T
LV
NZ
S
LT
N
CDN
SLO
BG
TPE
B
FIN
GR
CY
MEX
1993
I
RC
TPE
USA
I
GUS
H
D
CDN
SGP
CZ
A
RO
P
NZ
ROK
LV
IR
DK
AUS
NL
LT
SK
F
C
GB
T
BG
B
S
FIN
SLO
GR
MEX
N
CH
YV
CY
KWT
1994
N
RC
GB
USA
A
SGP
ROK
TPE
CZ
GUS
IR
D
H
RO
DK
I
T
NZ
UA
AUS
F
PL
NL
SK
CDN
LT
S
N
BG
FIN
EST
LV
CH
MEX
SLO
B
CY
GR
TR
YV
C
KWT
1995
RC
RC
IR
RO
A
D
GB
SK
TPE
I
CZ
RUS
H
AUS
SGP
F
TR
PL
USA
DK
RA
ROK
UA
LT
T
NL
CH
BG
S
NZ
EST
CDN
MEX
N
SLO
LV
CY
B
GR
FIN
YV
KWT
C
1996
RUS
IR
RC
RUS
A
D
USA
UA
CZ
H
RO
GB
TPE
BY
SGP
RA
TR
F
I
AUS
ROK
EST
CDN
T
VN
SK
CH
NL
NZ
DK
PL
SLO
MEX
LV
N
CY
BG
S
LT
E
B
GR
FIN
YV
C
KWT
1997
CDN
H
D
TR
TPE
IR
RUS
ROK
RC
SGP
PL
USA
UA
AUS
CDN
RO
A
T
EST
CZ
VN
F
S
BY
NZ
LV
RA
SLO
GB
SK
LT
I
DK
NL
IRL
N
MEX
CH
CY
E
FIN
BG
YV
GR
B
RI
KWT
C
1998
AUS
SGP
USA
ROK
RC
H
RA
RUS
AUS
D
GB
PL
A
RO
TPE
SK
NL
IR
UA
VN
LT
TR
BY
F
I
T
FIN
CZ
CDN
S
BG
N
MEX
CH
SLO
EST
CY
LV
DK
NZ
GR
KZ
E
IRL
B
KS
YV
RI
1999
T
USA
ROK
RC
IR
RO
H
TPE
UA
PL
AUS
VN
D
RA
BY
T
F
TR
SGP
IND
GB
RUS
MEX
A
IRL
NZ
I
CDN
LT
NL
SK
BG
KZ
DK
CH
CZ
FIN
B
S
CY
EST
LV
SLO
YV
BR
E
N
RI
GR
ROU
C
2000
DK
RC
RUS
USA
H
TPE
A
SK
BY
VN
TR
SGP
D
ROK
IR
CZ
FIN
T
MEX
GB
AUS
IND
CDN
RA
UA
PL
NZ
BG
F
DK
NL
B
RO
KZ
LT
CH
SLO
EST
S
YV
CY
HR
I
RI
N
AZ
IRL
E
LV
GR
BR
(List of abbreviations see page 132)
129
About the History of the IChO
IChO held in
1
.
.
.
5
.
.
.
.
10
.
.
.
.
15
.
.
.
.
20
.
.
.
.
25
.
.
.
.
30
.
.
.
.
35
.
.
.
.
40
.
.
.
.
45
.
.
.
.
50
2001
IND
RC
ROK
USA
RUS
IR
TR
IND
AUS
TPE
T
SGP
PL
RO
F
SK
H
VN
CZ
RA
BY
C
D
GB
UA
A
MEX
DK
CDN
EST
RI
HR
I
N
BG
CY
KZ
B
LT
NZ
CH
E
FIN
SLO
NL
LV
BR
S
YV
IRL
GR
2002
NL
RC
T
TPE
ROK
A
UA
USA
PL
IND
D
IR
H
RUS
CDN
TR
AUS
GB
SGP
E
SK
BY
VN
FIN
F
LT
CZ
KZ
LV
NL
RO
RA
EST
HR
BG
NZ
I
DK
SLO
N
YV
MEX
BR
S
RI
TM
B
IRL
CH
C
CY
2003
GR
RC
IR
ROK
T
BY
RUS
IND
SGP
D
TPE
UA
PL
CDN
CZ
RO
KZ
VN
EST
GB
AUS
H
SK
USA
YV
IND
F
A
I
TR
AZ
MEX
LT
NL
FIN
HR
J
DK
RA
GR
LT
E
TM
BR
BG
CH
NZ
IS
IRL
CY
KS
(List of abbreviations see page 132)
130
2004
D
RC
ROK
RUS
UA
D
PL
TPE
H
TR
VN
IND
IR
RO
LT
CZ
USA
SGP
CDN
AZ
AUS
KZ
GB
J
A
BY
SK
T
RA
EST
F
NZ
SLO
HR
LV
NL
I
CH
FIN
RI
S
BG
KS
E
GR
BR
TM
CY
YVA
IRL
IS
2005
TPE
ROK
VN
IR
RUS
AZ
TPE
T
RA
D
IND
A
CZ
UA
PL
AUS
TR
H
SK
USA
GB
RO
BY
SGP
J
RI
LV
BG
HR
MEX
KZ
LT
F
EST
CDN
I
DK
SLO
FIN
NL
IRL
GR
NZ
KS
S
B
BR
CH
P
IS
N
2006
ROK
RC
TPE
ROK
RUS
VN
T
J
Pl
IND
D
SK
DK
SGP
BR
CDN
AZ
UA
USA
H
CZ
AUS
IRL
F
IR
A
TR
RI
GB
RO
NL
HR
LT
KZ
SLO
EST
RA
BR
TJ
LV
MAL
S
IRL
IL
FIN
IS
I
CY
N
TM
CH
2007
RUS
RC
RUS
TPE
PL
ROK
D
T
IND
H
SK
LT
USA
VN
GB
BY
EST
UA
RI
IR
RO
AUS
A
KZ
SGP
NZ
CZ
F
TR
J
ARM
SLO
RA
BR
CDN
I
MAL
IL
IRL
NL
CH
S
LV
DK
MD
E
BG
TM
HR
PK
N
2008
H
RC
RUS
UA
ROK
T
BY
VN
TPE
H
SGP
KZ
A
PL
IR
IND
RO
AUS
D
SK
TR
LT
EST
I
GB
CDN
NZ
BR
USA
LV
RI
F
CZ
J
DK
RA
MEX
SLO
IL
AZ
HR
TM
BG
MGL
IRL
MAL
E
S
NL
CH
ROU
2009
GB
TPE
RC
ROK
RUS
SGP
J
USA
H
IR
GB
RO
T
D
IND
PL
AUS
A
BY
VN
F
RI
TR
LT
UA
EST
CZ
SK
CDN
I
RA
NZ
TM
MEX
KZ
IL
BR
HR
AZ
DK
S
LV
IRL
FIN
N
E
NL
MGL
PE
PK
SLO
2010 2011
J
TR
RC
RC
T
ROK
ROK RUS
J
RI
TPE USA
H
T
CZ
SGP
SGP CDN
USA
H
IR
IR
RUS
TR
TR
IND
LT
CZ
D
F
PL
J
GB
TPE
IND
D
RI
SK
RO
KZ
A
AUS
VN
VN
SK
RO
CDN
GB
EST
BY
AUS
PL
UA
A
F
LT
RA
EST
NZ
RA
BY
UA
KZ
FIN
BR
SLO
IL
I
HR
BR
SLO
HR
FIN
NZ
DK
TM
NL
LV
E
S
I
NL
LV
PE
BG
PK
CR
TJ
CH
E
IRL MEX
MEX CH
MGL MGL
MAL
IL
N
CY
S
BG
2012
USA
TPE
ROK
RUS
IND
RC
SGP
J
D
H
UA
RI
USA
BY
VN
RO
LIT
CZ
KZ
RA
PL
SK
IR
A
GB
AUS
IL
HR
BR
CDN
NZ
TR
EST
LV
F
ARM
I
NL
TM
DK
TJ
YVA
BG
SLO
CH
FIN
MEX
MGL
T
PK
AZ
About the history of the IChO
2013
IChO held in RUS
1
RC
.
ROK
.
TPE
.
USA
5
H
.
SGP
.
RUS
.
PL
.
UA
10
IND
.
VN
.
T
.
BY
.
J
15
KZ
.
IR
.
SK
.
CZ
.
RI
20
D
.
RO
.
A
.
LIT
.
AUS
25
GB
.
TR
.
NZ
.
HR
.
F
30
DK
.
MD
.
CDN
.
LV
.
SLO
35
RA
.
SRB
.
BR
.
EST
.
UZ
40
AZ
.
I
.
E
.
IL
.
CY
45
N
.
ARM
.
PK
.
CH
.
BG
50
TJ
2014
VN
SGP
UA
RUS
VN
TPE
RC
USA
TR
RO
T
IR
PL
ROK
RI
J
BY
GB
D
LT
IND
SK
CZ
H
AUS
UZ
CDN
SRB
RA
MEX
A
NZ
EST
KZ
MAL
KSA
HR
DK
BR
NL
PK
F
I
BG
E
SLO
TM
LV
CH
PE
N
2015
AZ
RC
ROK
TPE
SGP
RO
RUS
J
IND
USA
PL
TR
UA
T
KZ
IR
CS
VN
BY
SK
GB
SRB
A
LT
H
EST
CDN
RI
D
LV
I
RA
AUS
BR
BG
MRL
PE
DK
KSA
CH
MD
F
NZ
IL
UZ
SLO
PK
FIN
AZ
KS
NL
2016
2017
2018
2019
2020
2021
2022
2023
2024
(List of abbreviations see page 132)
131
About the History of the IChO
List of abbreviations
A
ARM
AUS
AZ
B
BG
BR
BY
C
CDN
CH
CS
CY
CZ
D
DDR
DK
E
EAK
EST
ET
F
FIN
GB
GR
GUS
H
HR
I
IL
IND
IR
IRL
IS
J
KS
KSA
KWT
KZ
132
Austria
Armenia
Australia
Azerbaijan
Belgium
Bulgaria
Brazil
Belarus
Cuba
Canada
Switzerland
Czechoslovakia
Cyprus Republic
Czech Republic
Germany
German Democratic Republic
Denmark
Spain
Kenya
Estonia
Egypt
France
Finland
United Kingdom
Greece
Commonwealth of Independent States
Hungary
Croatia
Italy
Israel
India
Iran
Ireland
Iceland
Japan
Kyrgyzstan
Saudi Arabia
Kuwait
Kazakhstan
LV
LT
MAL
MD
MEX
MGL
N
NL
NZ
P
PE
PK
PL
RA
RI
RC
RO
ROK
ROU
RUS
S
SGP
SK
SLO
SRB
SU
T
TJ
TM
TPE
TR
UA
USA
UZ
VN
WAN
YU
YV
Latvia
Lithuania
Malaysia
Moldova
Mexico
Mongolia
Norway
Netherlands
New Zealand
Portugal
Peru
Pakistan
Poland
Argentina
Indonesia
China
Romania
South Korea
Uruguay
Russian Federation
Sweden
Singapore
Slovakia
Slovenia
Serbia
Soviet Union
Thailand
Tadzhikistan
Turkmenistan
Chinese Taipei
Turkey
Ukraine
United States of America
Uzbekistan
Vietnam
Nigeria
Yugoslavia
Venezuela